This is a guess, or extended-comment, but I think no:
So in homology, $H_1$ is abelianization, $H_2$ is Schur multiplier, and $H_{n\ge 3}$ is ???
Likewise in cohomology, $H^1(G;A)$ is split extensions. $H^2(G,A)$ corresponds to group extensions $\mathcal{E}(G,A)$ of $G$ by $A$, and this makes its appearance in your UCT (assuming for simplicity that $G$ is abelian and $A$ is a trivial $G$-module); the map $H^2(G;A)\to Hom(H_2(G),A)$ tells us that every alternating map comes from a 2-cocycle, and we note in general that $Ext^1_R(M,N)$ is the set of $R$-module extensions of $M$ by $N$.
As we see, this all ties in to knowing $H^2(G;A)$... if we look at $H^3(G;A)$, we get crossed module extensions $0\to A\to N\to E\to G\to 0$, and these get rather cumbersome. We have no nice interpretation for $H^n(G;A)$ for $n>3$, except more crazy looking exact sequences.
This is why I don't expect a nice map ("interpretation" of it) in the UCT to arise.

This is an extended-comment, but I think no:

In homology, we know that $H_1$ is abelianization, $H_2$ is Schur multiplier, but $H_{n\ge 3}$ is ???
In cohomology, $H^1(G;A)$ is split extensions, and this fits well with $Hom(G,A)$ in the UCT. As you mention, $H^2(G,A)$ corresponds to group extensions of $G$ by $A$, and this fits well with $Hom(H_2G,A)$ in the UCT. But as we see, this all ties in to knowing $H^n(G;A)$ and $H_n(G;A)$ very well...

If we look at $H^3(G;A)$ we get crossed module extensions $0\to A\to N\to E\to G\to 0$, and these are cumbersome. We have no nice interpretation for $H^n(G;A)$ for $n>3$, except more crazy-looking exact sequences. This is why I don't expect a "nice" map (i.e. "interpretation" of it) in the UCT to arise.