You're only finding UCT in the literature for trivial group actions, *because there is no general UCT for nontrivial group actions*:

The general Kunneth formula does not hold for arbitrary groups and actions. But it does hold sometimes, and I elaborated on this here: http://mathoverflow.net/questions/75472/kuenneth-formula-for-group-cohomology-with-nontrivial-action-on-the-coefficient/75485#75485.

Now the UCT, which relates $H_*(G,M)$ to $H_*(G,\mathbb{Z})$, only follows from the Kunneth formula **for trivial group actions**.  The Kunneth theorem considers the tensor product $C_*\otimes D_*$ of two chain complexes, and the special case for UCT is $D_*=M$ for some $R$-module... to guarantee that the images of the boundary maps are $R$-projective, some extra assumptions are needed (like $R$ is a PID).  For group homology, we work with $F_*\otimes_{\mathbb{Z}G}M$ where $F_*$ is a free resolution of $\mathbb{Z}$ as a $\mathbb{Z}G$-module.  We cannot take $R=\mathbb{Z}G$ (otherwise the assumption about the boundaries would imply that all homology groups are trivial), so we take $R=\mathbb{Z}$.  But then $M$ must be trivial as a $\mathbb{Z}G$-module in order to express $F_*\otimes_{\mathbb{Z}G}M$ as $C_*\otimes_\mathbb{Z}M$.  In this case, $F_*\otimes_{\mathbb{Z}G}M=(F_*\otimes_{\mathbb{Z}G}\mathbb{Z})\otimes_\mathbb{Z}M$ and we can apply the UCT.