Take a finite group $G$, a finite-dimensional $\mathbb{Q}$-vector space $V$ and two representations actions of $G$ that are inequivalent but the spaces of coinvariants both have the same dimension, say zero.
The first concrete example that comes to my mind is $G=Z/4$, $V=Q[i]$, with the two actions where the generator acts by $-1$ or $i$. These two actions are not even equivalent under outer automorphisms of $G$.
Let $n \geq 3$ be an odd integer and consider the Eilenberg-Mac-Lane space $K(V,n)$, which inherits two $G$-actions. The two Borel-constructions $EG \times_G K(V,n)$ have the same homotopy groups. But the $n$th homotopy groups are not isomorphic when considered as a $\pi_1$-module; so these two spaces are not homotopy equivalent.
The homology can be computed from the Leray-Serre spectral sequence of the fibration $EG \times_G K(V,n) \to BG$. Recall $E^{2}_{pq}= H_p (G; H_p (K(V;n)))$.
To begin with, $\tilde{H}_* (K(V,n), \mathbb{Z}) =V$ if $*=n$ and $0$ otherwise. Thus $E^{2}_{pq}=0$ unless $q=0$ (then it is the group homology $H_p(G;Z)$ or $(p,q)=(0,n)$, in which case it is $V_G=0$. Thus the projections $EG \times K(V,n) \to BG$ are homology equivalences.
Finally note that the Eilenberg Mac Lane spaces can be realized as abelian topological groups and $G$ acts fixing the basepoint. Thus the maps $EG \times_G K(V;n)) \to BG$ have sections, which are homology equivalences as well. So the construction even produces a homology equivalence between two spaces with abstractly isomorphic homotopy groups.
EDIT: since there is a homology equivalence between these two spaces, it follows that the homology with coefficients, the cohomology rings and even the actions of the Steenrod algebra for all primes are isomorphic.