I wouldn't use MSE simply because $\hat{F}$ doesn't have constant variance. If you used a weighted MSE that should do better ... but then $\hat{F}_i$ and $\hat{F}_j$ aren't independent, either.
One possibility if you're trying to make $\hat F$ as close to $F$ in a MSE-like sense as possible is you might try minimizing the Anderson-Darling statistic (since it's like a precision-weighted MSE).
To be honest, I'd be inclined to use likelihood to estimate the parameters, if at all feasible -- but not on the ECDF; you want to deal with the data and construct the likelihood (i.e. look at data values and $f$, not $F$ and $\hat{F}$) if you difference back to what's essentially a histogram$^\dagger$ you should be able to get good ML estimates* from that.

$\hspace{3cm}\to$ 4 5 5 5 6 6 6 7 7 7 7 8 9 9 9 9 9 9 ...
* (if slightly approximate because of the discretization)
$\dagger$ - In case it's not obvious how one obtains the sample size in the case where you only have an ecdf and not $n$, let me be explicit about it. The situation you seem to be dealing with (i.e. as "at a suitably fine grid" suggests) we have a finely discretized continuous distribution, one we might reasonably still treat as if it were continuous if only we had the observations.
Consider that at each jump point, the ecdf must increase by some multiple of $1/n$.
The chance that it will always increase by a multiple of $k/n$ ($2/n$ or $3/n$ say) at every jump point would be extremely small indeed (some simple hand calculations or simulations make the point well enough). So we can - with probability extremely close to 1 - infer the exact $n$ by finding the largest $1/n$ consistent with every change in ecdf.
[On the other hand, with a very small sample at a very coarse grid, so that only a few different values will occur, you could get the increase at each point being $2/n$ or perhaps $3/n$ and so run into problems, particularly if you want standard errors. So if we only had 4 bins and the counts in those bins were 2,6,12,4, we'd get $n$ wrong (it would look like the jumps were multiples of 1/12 rather than 1/24, so we'd think the $n_i$ were 1,3,6,2), and so the standard errors - and in some cases, even the point estimates - could be a fair way off. Fortunately, this isn't the case for this question.]