Let $\Sigma$ be a surface (let's say oriented and of finite type). We can consider the configuration space $F(\Sigma,n)$ of $n$ ordered distinct points on $\Sigma$, i.e. $\Sigma^n\setminus \Delta$ where $\Delta$ is the "big diagonal". The cohomology of $F(\Sigma,n)$ can be computed in a very explicit way, as explained in Burt Totaro's paper "Configuration spaces of algebraic varieties": the Leray spectral sequence for $F(\Sigma,n) \hookrightarrow \Sigma^n$ degenerates after the first differential and can be written down in a concrete way, which gives us a completely explicit differential graded algebra whose cohomology is $H^\bullet(F(\Sigma,n))$.

Now fix a finite group $G$ and consider the space $F(\Sigma,G,n)$ which parametrizes $n$ points on $\Sigma$ and a principal $G$-bundle over the complement of the $n$ points on $\Sigma$. So we have a finite sheeted covering $F(\Sigma,G,n) \to F(\Sigma,n)$ such that the fiber over the point $(s_1,\ldots,s_n) \in \Sigma^n$ is the set $\mathrm{Hom}(\Pi_1(\Sigma \setminus \{s_1,\ldots,s_n\}),G)/G$, where $G$ acts on the set of maps by conjugation.

Q1. Is there a good way to describe or compute the cohomology of $F(\Sigma,G,n)$?

Q2. (A vaguer question.) Let's say $\Sigma$ is compact for simplicity and let $FM(\Sigma,n)$ be the Fulton-MacPherson compactification of $F(\Sigma,n)$. I think there is a natural compactification $$F(\Sigma,G,n) \hookrightarrow FM(\Sigma,G,n)$$ covering $F(\Sigma,n) \hookrightarrow FM(\Sigma,n)$ and such that the covering $FM(\Sigma,G,n) \to FM(\Sigma,n)$ ramifies along the boundary; $FM(\Sigma,G,n)$ should parametrize principal $G$-bundles which are allowed to ramify over the nodes. Is there a natural smaller compactification of $F(\Sigma,G,n)$ which is still smooth, analogous to the compactification $$F(\Sigma,n) \hookrightarrow \Sigma^n$$ (which of course is smaller than the Fulton-MacPherson)?

Craig Westerland suggests an alternative description of the cohomology of $F(\Sigma,G,n)$. The covering $p \colon F(\Sigma,G,n) \to F(\Sigma,n)$ satisfies $R^ip_\ast\mathbf Z =0$ for $i>0$, so $$ H^\bullet(F(\Sigma,G,n),\mathbf Z) = H^\bullet(F(\Sigma,n),p_\ast\mathbf Z). $$ Now $F(\Sigma,n)$ is a $K(\pi,1)$ space where $\pi$ is by definition the pure braid group on $n$ strands of the surface $\Sigma$, $P_n(\Sigma)$. Hence this cohomology is given by $$ H^\bullet(P_n(\Sigma), \mathbf Z [ \mathrm{hom}(\pi_1(\Sigma \setminus \{s_1,\ldots,s_n\},G)/G]).$$ Also the action of $P_n(\Sigma)$ is the restriction of an action of the $n$-strand surface braid group $B_n(\Sigma)$ (i.e. where the points are unordered).

The simplest example should be $\Sigma = \mathbf R^2$, where we get the usual pure braid group. Since the fundamental group of the punctured plane is free we find that $\mathrm{hom}(\pi_1(\Sigma \setminus \{s_1,\ldots,s_n\},G)/G = G^n/G$, where $G$ acts by elementwise conjugation on $G^n$. The action of the braid group $B_n$ can be written down in terms of Artin's generators $\sigma_i$: the element $\sigma_i$ acts by $$ (g_1,\ldots,g_i,g_{i+1},\ldots,g_n) \mapsto (g_1,\ldots,g_ig_{i+1}g_i^{-1},g_i,\ldots,g_n)$$ (which is well defined on equivalence classes modulo conjugation). Then I guess even these cohomology groups are hard to compute in general?