Sorry, this is my fault. $S^n(G)$ is not an algebraic group and one has only the map of schemes $S^n(G)\to G$. Nevertheless, $S^\bullet(G)$ is a commutative monoidal scheme and the question is still valid

Dear @JacksonMorrow, thank you for the paper but I think that Cor. 3.6 states something else. Roughly speaking, it says that the functor of taking the underlying topological spaces of the analytification commutes with a finite group quoitent. This does not help with contractibility.

Basically I can assume that the singular locus of $X/G$ is smooth, i.e $X/G$ has isolated singularities. But I guess that the fact is true without this assumption.