Does Frisch–Waugh–Lovell theorem change the covariance matrix on regression coefficients? I know FWL theorem allows you to compute a subset of the regression coefficients, i.e. for fixed effects models.
It can give you identical regression coefficients as you solve the whole regression problem.
But does this change the covariance matrix of the coefficients?
 A: I am not sure what you precisely mean by "change", but there is indeed a pitfall. 
When you do fixed effects, you can do what you call the "whole regression problem", i.e., regress $y_{it}$ on $X_{itk}$, $k=1,\ldots,K$, $i=1,\ldots,n$, $t=1,\ldots,T$ and $n$ individual-specific intercepts, for a total of $n+K$ coefficient estimates.
Alternatively, you can do FWL, which amounts to first subtracting the individual-specific averages from all the $y_i$ and $X_{it}$ tp get $y_i-\bar{y}_i$ and $X_{itk}-\bar{X}_{ik}$ and then regress $y_i-\bar{y}_i$ on $X_{itk}-\bar{X}_{ik}$. This second regession only has $K$ regressors. 
If you now (under homoskedasticity) estimate the error variance $\sigma^2$, a component of the variance-covariance matrix, by the usual formula $SSR/(n\cdot T-K)$ you make a mistake, because the formula neglects the fact that you have already fitted $n$ coefficients in the first step. That mistake moreover does not vanish even asymptotically, as $n$ of course keeps growing if you add more individuals to the panel.
Instead, $SSR/(n\cdot T-n-K)$ can be shown to converge to $\sigma^2$.
