I have described the gist of McNemar's test rather extensively here and here, it may help you to read those. Briefly, McNemar's test assesses the balance of the off-diagonal counts. If people were as likely to transition from approval to disapproval as from disapproval to approval, then the off-diagonal values should be approximately the same. The question then is how to test that they are. Assuming a 2x2 table with the cells labeled "a", "b", "c", "d" (from left to right, from top to bottom), the actual test McNemar came up with is:
$$
Q_{\chi^2} = \frac{(b-c)^2}{(b+c)}
$$
The test statistic, which I've called $Q_{\chi^2}$ here, is approximately distributed as $\chi^2_1$, but not quite, especially with smaller counts. The approximation can be improved using a 'continuity correction':
$$
Q_{\chi^2c} = \frac{(|b-c|-1)^2}{(b+c)}
$$
This will work better, and realistically, it should be considered fine, but it can't be quite right. That's because the test statistic will necessarily have a discrete sampling distribution, as counts are necessarily discrete, but the chi-squared distribution is continuous (cf., Comparing and contrasting, p-values, significance levels and type I error).
Presumably, McNemar went with the above version due to the computational limitations of his time. Tables of critical chi-squared values were to be had, but computers weren't. Nonetheless, the actual relationship at issue can be perfectly modeled as a binomial:
$$
Q_b = \frac{b}{b+c}
$$
This can be tested via a two-tailed test, a one-tailed 'greater than' version, or a one-tailed 'less than' version in a very straightforward way. Each of those will be an exact test.
With smaller counts, the two-tailed binomial version and McNemar's version that compares the quotient to a chi-squared distribution, will differ slightly. 'At infinity', they should be the same.
The reason R cannot really offer a one-tailed version of the standard implementation of McNemar's test is that by its nature, chi-squared is essentially always a one-tailed test (cf., Is chi-squared always a one-sided test?).
If you really want the one-tailed version, you don't need any special package, it's straightforward to code from scratch:
Performance
# 2nd Survey
# 1st Survey Approve Disapprove
# Approve 794 150
# Disapprove 86 570
pbinom(q=(150-1), size=(86+150), prob=.5, lower.tail=FALSE)
# [1] 1.857968e-05
## or:
binom.test(x=150, n=(86+150), p=0.5, alternative="greater")
# Exact binomial test
#
# data: 150 and (86 + 150)
# number of successes = 150, number of trials = 236, p-value = 1.858e-05
# alternative hypothesis: true probability of success is greater than 0.5
# 95 percent confidence interval:
# 0.5808727 1.0000000
# sample estimates:
# probability of success
# 0.6355932
Edit:
@mkla25 pointed out (now deleted) that the original pbinom() call above was incorrect. (It has now been corrected; see revision history for original.) The binomial CDF is defined as the proportion $≤$ the specified value, so the complement is strictly $>$. To use the binomial CDF directly for a "greater than" test, you need to use $(x−1)$ to include the specified value. (To be explicit: this is not necessary to do for a "less than" test.) A simpler approach that wouldn't require you to remember this nuance would be to use binom.test(), which does that for you.
