The argument in the quoted paper is a bit too sketchy. An actual proof will have two parts:
Let $f:X \to Y$ be map. Assume that for each $y \in Y$, $\tilde{H}^{\ast} (f^{-1}(y);\mathbb{Q})=0$ (this is what the phrase ''$\mathbb{Q}$-acyclic'' means). Find conditions that guarantee that $f$ induces an isomorphism in cohomology. Essentially, one needs to provide assumptions that the Leray spectral sequence of the map is as nice as claimed in Marks answer.
Provide conditions for group action of a topological group $G$ on a space $X$ such that $EG \times_G X \to X/G$ has the properties found in 1 (and the answer in 1 depends on the example you want to consider).
ad 1. Just having finite stabilizers is not enough. Consider the action of $G=\mathbb{Z}$ on $X=S^1$ by an irrational rotation. Then $EG \times_G S^1 \simeq S^1 \times S^1$. The quotient space is an uncountable set with the trivial topology. Because each map into this space is continuous, it is contractible and has trivial (Cech or singular) cohomology.
One needs criteria that show that a map $f:X \to Y$ with contractible/acyclic/$\mathbb{Q}$-acyclic fibres is itself a weak equivalence/acyclic/$\mathbb{Q}$-acyclic. Having contractible fibres does not suffice: look at the identity map $[0,1]^{\delta} \to [0,1]$, where $\delta$ means ''discrete topology''.
Such problems are solved by what I would call ''fibre theorems''. You wish to know how close the natural map $f^{-1}(y) \to hofib_f (y)$ from the point-preimages to the homotopy fibres is to be a weak homotopy equivalence. What one needs is a suitable local condition and then a local-to-global result. Many technical results are of this type: ''fibre bundles are Serre-fibrations'', ''local Hurewicz fibrations are Hurewicz fibrations'', Dold-Thom's result on quasifibrations, the Vietoris-begle mapping theorem, Quillens Theorems A and B and McDuff-Segals formulation of the ''group-completion'' theorem. A variation of the latter solves your problem:
Assume, as above, that for each $y \in Y$, $f^{-1}(y)$ is $\mathbb{Q}$-acyclic.
Assume moreover that $Y$ is Hausdorff and for each $y \in U \subset Y$, there is $y \in V \subset U$ ($U$ and $V$ open) such that $V$ is contractible and $ f^{-1} (V)$ is $\mathbb{Q}$-acyclic (let us call these sets ''good''). Then, I claim, $f$ induces an isomorphism in rational cohomology.
Proof: Let $P:=\coprod_U U$, where $U$ runs through all good sets. This is a topological poset: $(U,x) \leq (V,y)$ iff $x=y$ and $U \subset V$. Take the nerve $|N_{\bullet} P|$ and consider the map $g:|N_{\bullet} P| \to Y$. It has contractible point-preimages: the preimage of $x$ is the poset of all good sets containing $x$, ordered by inclusion and the assumption says that this has contractible realization. In their recent paper arXiv:1201.3527, Galatius and Randal-Williams prove a lemma that under the present assumptions, $g$ is a Serre fibration with contractible fibres, in particular a weak homotopy equivalence.
Apply the same argument to get a poset $Q$, formed not from good subsets of $Y$, but their preimages in $X$. Thus up to weak equivalence, we can replace $f:X \to Y$ by $|N_{\bullet} Q| \to |N_{\bullet} P|$. By assumption, on the $p$th simplicial stage, the map $N_p Q \to N_p P$ is a rational homology equivalence (because this is just a disjoint union of the maps $f^{-1}(U) \to U$). The spectral sequence of simplicial spaces gives then that $|N_{\bullet} Q| \to |N_{\bullet} P|$ is a rational homology equivalence.
ad 2. There are several instances where the above argument can be applied. Example 1: let $G$ be a discrete group that acts simplicially on a simplicial complex, with finite stabilizers.
Example 2: let $G$ be a Lie group that acts properly on a smooth manifold with finite stabilizers. By some standard results, there is a $G$-invariant Riemannian metric and $G$-equivariant tubular neighborhoods of orbits $Gx\cong G/G_x$. The image of such a tubular neighborhood $U \subset M$ is a contractible subset of $M/G$. The preimage of $U/G$ under $EG \times_G M \to M/G$ is $EG \times_G U \simeq EG \times_G G/G_x \simeq BG_x$, which is $\mathbb{Q}$-acyclic.
Best Answer
These results follow from the Cartan-Leray spectral sequence, which for a regular covering map $X\to X/W$ and a commutative ring $k$ of coefficients has $$ E_2^{p,q}=H^p(W,H^q(X;k)) $$ (cohomology of the group $W$ with coefficients in the $kW$-module $H^\ast(X,k)$) and converges to a graded group associated to $H^\ast(X/W)$. A reference is Ken Brown's "Cohomology of groups", section VII.7.
In case the group $W$ is finite, if $|W|$ is invertible in $k$ then $H^p(W;H^q(X;k))=0$ for all $q$ and all $p>0$ (see Brown, Corollary III.10.2). In particular this is true if $k=\mathbb{Q}$. Thus the spectral sequence is concentrated in the $0$ column and therefore collapses, giving $H^\ast(X/W)\cong H^0(W;H^\ast(X;k))$. Since $H^0(W;M)=M^W$ for any group $W$ and any $W$-module $M$, this gives the stated results. So you were exactly right in your first paragraph!
More generally, the same collapse happens for the Serre spectral sequence of the fibration $X\to X_W\to BW$, which has $$ E_2^{p,q}=H^p(BW;H^q(X))\cong H^p(W;H^q(X)), $$ giving the isomorphism $H^\ast_W(X)\cong H^\ast(X)^W$ you mentioned.