For ease of referencing I'll prove the stated isomorphism under the hypothesis that $G$ is a compact Lie group and $X$ is a finite dimensional compact topological $G$-manifold.
By the hypothesis on $G$ we can choose $EG$ to be a dimension-wise finite CW complex. The $(n+1)$-skeleton $E$ is compact, $n$-connected and for fixed $m$ and $n$ large enough, the inclusion $E \hookrightarrow EG$ induces an isomorphism $$H^m(EG\times_G X,\mathbb{Q}) \xrightarrow{\cong} H^m(E\times_G X,\mathbb{Q})$$ (that's 4) on p. 4 of the linked paper in the question). Hence it suffices to show that the map $E\times_G X \to X/G$ induces an isomorphism in rational cohomology in degree $m$.
But this follows from the Vietoris-Begle theorem [Q, Corollary A.7] that applies if we have shown that $f$ is closed and the fibres of $f$ are compact, n-acyclic and relative Hausdorff in $E\times_G X$:
Closedness of $f$ and compactness of the fibres $E/G_x$ are obvious, $n$-acyclity follows from $H^i(E/G_x,\mathbb{Q})=H^i(G_x,\mathbb{Q})=0$ for $0< i< n$. Relative Hausdorff means different points in the fibre have disjoint neighborhoods in $E \times_G X$. This holds since $E \times_G X$ is Hausorff. QED
[Q] Quillen: The Spectrum of an Equivariant Cohomology ring: I
The cited Vietoris-Begle theorem states:
If $f: X \to Y$ is closed and its fibres are compact, relatively Hausdorff in $X$ and $n$-acyclic, i.e. $H^i(X,k)=H^i(\ast,k)$$H^i(f^{-1}(y),k)=H^i(\ast,k)$ for $i< n$ ($k$ any constant coefficients), then $f^\ast:H^i(Y,k) \xrightarrow{\cong} H^i(X,k)$ for $i \lt n$.