What is $BHom(G;H)$? Typically, $Hom(G;H)$ is not a group unless $H$ is abelian. Maybe you want to talk about the natural map

$$
Hom(G;H) \to Map_0 (BG;BH)
$$

from the space of homomorphisms to the mapping space. In some cases, this is a homotopy equivalence, for example if $G$ is connected and compact and $H=U(1)$. In that case, $Hom(G,H)$ is discrete (it is $\cong Hom(\pi_1(G),\mathbb{Z})$), and $Map_0 (BG;BU(1))=Map_0 (BG;K(\mathbb{Z};2))$. The latter space has homotopy groups $\pi_i (Map_0 (BG;K(\mathbb{Z};2))) = \tilde{H}^{2-i} (BG;\mathbb{Z})$, which is zero unless $i=0$ and it is not so difficult to see that $Hom (G,H)\to Map_0 (BG;BU(1))$ is a bijection on components.

If the target $H$ is not abelian, things become much more difficult. An example is $Hom(S^3;S^3)$. Up to conjugacy, there are only two homomorphisms $S^3 \to S^3$. But there are infinitely many homotopy classes of maps $BS^3 \to BS^3$. This was proven by Sullivan.
One can show that the degree of a map $BS^3 \to BS^3$ (meaning the effect on $H^4 (BS^3)=\mathbb{Z}$) is always an odd square. This is discussed in Hatchers textbook. Sullivan then proved, using profinite homotopy theory, that for each odd square, there is indeed a map with that degree.

As far as I remember, questions of this kind are also discussed in the series "maps between classifying spaces" by Adams and Mahmud, but I do not know enough to say something on it.

EDIT: if $H$ is assumed to be discrete, then the map is always a weak homotopy equivalence.
Proof: In that case, $Hom(G;H)=Hom(\pi_0 (G);H)$, with the discrete topology. Moreover, for each space $X$ (which is homotopy equivalent to a CW complex), there is a homotopy equivalence $Map_0(X;BH)\cong Hom(\pi_1 (X);H)$; insert $X=BG$.

EDIT2: if $G$ is discrete, but infinite, then almost nothing can be said. Look at $H=U(1)$ and let $G$ be a perfect group. Then each homomorphism $G \to U(1)$ is trivial, but $Map(BG,BU(1))$ has the homotopy type $H^2 (G)$ (its components are contractible).