Can McNemar's test be improved upon by adjustments for zeros like those in a sign test? McNemar's test is a special example of the binomial sign test, but the ('vanilla') sign test suffers from bias due to ignoring differences equal to zero.
McNemar's test statistic is given by:
$\chi^{2} = \frac{\left(|r-s|-1\right)^{2}}{r+s}$, where $r$ and $s$ are the counts of discordant pairs (0,1) versus (1,0), distributed $\chi^{2}$ with 1 degree of freedom under the null hypothesis.
I am having a hard time parsing Sribney on the sign test:

The test statistic for the sign test is the number $n_{+}$ of observations greater than zero. Assuming that the probability of an observation being equal to zero is exactly zero, then, under the null hypothesis, $n_{+} \sim \text{Binomial}(n, p=\frac{1}{2})$, where $n$ is the total number of observations. But what do we do if we have some observations that are zero?
Fisher’s Principle of Randomization
We have a ready answer to this question if we view the test from the perspective of Fisher’s Principle of Randomization (Fisher 1935). Fisher’s idea (stated in a modern way) was to look at a family of transformations of the observed data such that the a priori likelihood (under the null hypothesis) of the transformed data is the same as the likelihood of the observed data. The distribution of the test statistic is then produced by calculating its value for each of the transformed “randomization” data sets, considering each data set equally likely.
For the sign test, the “data” are simply the set of signs of the observations. Under the null hypothesis of the sign test, $P(X_{i}>0)= P(X_{i}<0)$, so we can transform the observed signs by flipping any number of them and the set of signs will have the same likelihood. The $2^{n}$ possible sign changes form the family of randomization data sets. If we have no zeros, this procedure again leads to $n_{+} \sim \text{Binomial}(n, p=\frac{1}{2})$.
If we do have zeros, changing their signs leaves them as zeros. So if we observe $n_{0}$ zeros, each of the $2^{n}$ sign-change data sets will also have $n_{0}$ zeros. Hence, the values of $n_{+}$ calculated over the sign-change data sets range from 0 to $n-n_{0}$, and the “randomization” distribution of $n_{+}$ is $n_{+} \sim \text{Binomial}(n-n_{0}, p=\frac{1}{2})$.

Because this seems to be saying go ahead and ignore zeros. But then later in the paper, Sribney provides an adjustment for the sign-rank test that accounts for zeros just along the lines I am wondering about:

The adjustment for zeros is the change in the variance when the ranks for the zeros are signed to make $r_{j}=0$; i.e., the variance is reduced by $\frac{1}{4}\sum_{i=1}^{n-{0}}{i^{2}}=n_{0}\frac{\left(n_{0}+1\right)\left(2n_{0}+1\right)}{24}$.

Should I instead be asking whether or not to apply the signed-rank test to individually-matched case control data?
A simple made up example will illustrate why ignoring zeros presents a problem. Imagine you've paired data with no differences equal to zero (this would correspond to data for a McNemar's test with only discordant pairs present). With a sample size of, say, 20, you find 15 positive signs of differences and 5 negative signs of differences, and conclude significant difference. Now imagine that you have 1000 observed differences equal to zero in addition to those 15 positive and 5 negative signs of differences: now you conclude difference is not significant. If McNemar's test is conducted on 1020 pairs, 1000 of which are zeros, and with discordant pairs of 15 and 5, we should not reject the null hypothesis (e.g. at $\alpha = 0.05$).
There is an adjustment to the sign test to correct for observed zero differences based upon Fisher’s "Principle of Randomization" (Sribney, 1995).
Is there a way of improving on McNemar's test that addresses the effect of observed zero differences (i.e. by accounting for number of concordant pairs relative to number of discordant pairs)? How? What about for the asymptotic z approximation for the sign test?

References
Sribney WM. (1995) Correcting for ties and zeros in sign and rank tests. Stata Technical Bulletin. 26:2–4.
 A: E.L. Lehmann, J.P. Romano. Testing Statistical Hypotheses. 3rd ed. Springer, 2005. P. 136:

P(-), P(+) and P(0) denote the probabilities of preference for product
  [A over B, B over A, or A=B, tie], ... The hypothesis to be tested H0: P(+)=P(-)
  ... The problem reduces to that of testing the hypothesis P=1/2 in a
  binomial distribution with n-z [n = sample size, z = number of ties]
  trials ... The unbiased test is obtained therefore by disregarding the
  number of cases with no preference (ties), and applying the sign test
  to the remaining data.
The power of the test depends strongly on P(0) ... For large P(0), the
  number n-z of trials in the conditional [on z] binomial distribution
  can be expected to be small, and the test will thus have little power
  ... A sufficiently high value of P(0), regardless of the value of
  P(+)/P(-), implies that the population as a whole is largely
  indifferent with respect to the product.
As an alternative treatment of ties, it is sometimes proposed to
  assign each tie at random (with probability 1/2 each) to either plus
  or minus ... The hypothesis H0 becomes P(+)+1/2P(0)=1/2 ... This test
  can be viewed also as a randomized test ... and it is unbiased for
  testing H0 in its original form ... [But] Since the test involves
  randomization other than on the boundaries of the rejection region, it
  is less powerful than the [original test disregarding ties]..., so that the
  random breaking of ties results in a loss of power.

A: I don't see how this would be helpful, or even possible.  McNemar's test only uses the discordant pairs.  The Wikipedia page states:  

The McNemar test statistic is:
   $$
    \chi^2 = {(b-c)^2 \over b+c}. 
$$

I have a lengthy explanation of McNemar's test here: What is the difference between McNemar's test and the chi-squared test, and how do you know when to use each?  The whole post may be of value, but borrowing the example there:  
\begin{array}{rrrrrr}
            &            &{\rm After} &          & &             \\
            &            &{\rm No}    &{\rm Yes} & &{\rm total}  \\
{\rm Before}&{\rm No}    &1157        &35        & &1192         \\
            &{\rm Yes}   &220         &13        & &233          \\
            &            &            &          & &             \\
            &{\rm total} &1377        &48        & &1425         \\
\end{array}
McNemar's test is the binomial test of $220/(220+35)$; the concordant pairs (i.e., $1157$, and $13$) don't show up.  
