You can indeed! By Yoneda, *any* natural way of getting cohomology classes can be realized on the Eilenberg-MacLane spaces. You can look at those "freebie" classes in $H^n(K(\mathbb{Z}, n); G)$ coming from the universal class in $H^n(K(\mathbb{Z}, n); \mathbb{Z})$, and this will give your desired map $K(\mathbb{Z}, n) \to K(G, n)$. However, there's a nice bigger story behind this specific case besides just Yoneda.

First, the Dold-Kan theorem says that (nonnegatively graded) chain complexes of abelian groups are equivalent to simplicial abelian groups. More precisely, given a simplicial abelian group, the alternating sum of the face maps turns it into a chain complex, and modding out by the image of the degeneracies gives the "normalized" chain complex.  This turns out the be an equivalence of categories (the inverse is taking a chain complex and formally adding degenerate things to it). What's more, this equivalence preserves the usual notion of homotopy in the two categories: the derived category of abelian groups is the same as simplicial abelian groups with weak equivalences formally inverted.

Now $K(G, n)$ can be realized as the simplicial abelian group which corresponds to the chain complex with $G$ in degree $n$ and $0$ everywhere else. Maps of simplicial abelian groups (mod homotopy) from $K(G, n)$ to $K(H, m)$ are then the same as maps of chain complexes in the derived category, i.e. $\operatorname{Ext}^{m-n}(G, H)$. This is $0$ except for $m=n$ and $m=n+1$, and in those cases you get exactly the correspondence that the universal coefficient theorem gives you.  The cohomology operations with $m=n+1$ (coming from the $\operatorname{Ext}$ part of the universal coefficient theorem) are called Bocksteins and can also be obtained as the connecting homomorphisms of a long exact sequence on cohomology coming from a short exact sequence of coefficient groups (namely, the group extension corresponding to your element of $\operatorname{Ext}$).

Note that there are lots of other operations on cohomology (eg, Steenrod operations) besides these. What makes these special is that they can be implemented by maps from $K(G, n)$ to $K(H, m)$ which are *group homomorphisms* with respect to the abelian group structures on the spaces. Note, however, that any cohomology operation which commutes with addition (which includes all Steenrod operations) is a group homomorphism *up to homotopy*, since the group structure on cohomology is just the group structure on $K(G, n)$ taken modulo homotopy. Nevertheless, it is impossible to straighten these out to be homomorphisms on the nose, except in the case of Bocksteins and in the case $n=m$.