Is there any relation about rational homology of X and X/G

If we know the rational homology of X is 0, can we get some information about the rational homology of X/G, where G is a finite group? Thank you very much for the answers!

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When $G$ is finite, the rational cohomology of $X/G$ is the fixed point set $H^*(X;\mathbb{Q})^G$. This is proven in Grothendieck's Tohoku paper (Theorem 5.3.1 and the Corollary to Proposition 5.3.2).
So if the rational cohomology of $X$ is trivial, the same is true for $X/G$. And rationally the cohomology and homology are isomorphic.
For paracompact Hausdorff spaces, these cohomology groups can be taken to be the Cech cohomology groups. Note that if $X$ is homotopy equivalent to a CW complex, then Cech cohomology agrees with singular cohomology. You might also want to look at Oscar Randall-Williams comments here: http://mathoverflow.net/questions/18898/grothendiecks-tohoku-paper-and-combinatorial-topology/30015#30015.
@Agusti, this is a consequence of the existence of a spectral sequence $H^\bullet(G,H^\bullet(X,\mathbb Z))\Rightarrow H^\bullet(X/G,\mathbb Q)$. It is discussed (in a general form) in the sixth section of the Tôhoku paper, iirc (I don't have my copy at hand) It can readily be proved onces you know about the Grothendieck spectral sequence (once you know that sheaf cohomology agrees with singular cohomology on your nice space) – Mariano Suárez-Alvarez Sep 17 '10 at 13:55
is it generaly true for $X$ being a manifold or more, orbifold?(they only have homotopy type of CW complex)? – abc Sep 17 '10 at 14:05