Is the classifying space a fully faithful functor?

The functor $B : \mathbf{TopGrp} \to \operatorname{Ho} \mathbf{Top}_*$ is not fully faithful. Indeed, it does not even preserve non-isomorphy: the multiplicative group $\mathbb{R}^\times$ and the discrete group $\mathbb{Z} / 2 \mathbb{Z}$ have the same classifying space (namely, $\mathbb{R P}^\infty$), so if the functor were fully faithful, $\mathbb{R}^\times$ and $\mathbb{Z} / 2 \mathbb{Z}$ would have to be isomorphic as topological groups – but they are not.

The correct and invariant statement is that the classifying space functor is an equivalence of $(\infty, 1)$-categories between grouplike $E_1$ spaces (this is a homotopically invariant version of "topological group") and pointed connected spaces (and every time I say "spaces" I mean "weak homotopy types"). I believe it's also true that every grouplike $E_1$ space can be modeled by a topological group.

Trying to strictify this equivalence to a statement about homomorphisms beween topological groups is hopeless in general. The maps corresponding to maps $BG \to BH$ are maps $G \to H$ that respect the group operation up to coherent homotopy, so there's no reason to expect the classifying space functor, as a functor on topological groups and taking values in homotopy categories, to be full.

Explicitly, take $G$ to be discrete and take $H = S^1$. Then $[BG, BH] \cong H^2(G, \mathbb{Z})$, which by the universal coefficient theorem has two summands, namely

$$\text{Ext}^1(H_1(G, \mathbb{Z}), \mathbb{Z})$$

and $\text{Hom}(H_2(G, \mathbb{Z}), \mathbb{Z})$. Group homomorphisms $G \to S^1$ only give you terms in the first summand, via the Ext long exact sequence coming from the short exact sequence

$$0 \to \mathbb{Z} \to \mathbb{R} \to S^1 \to 0$$

giving a long exact sequence with connecting map $\text{Hom}(H_1(G), S^1) \to \text{Ext}^1(H_1(G, \mathbb{Z}), \mathbb{Z})$.

The second summand $\text{Hom}(H_2(G, \mathbb{Z}), \mathbb{Z})$ is related to the nontrivial structure implied by respecting group operations up to coherent homotopy. (This means that $f(g_1 g_2)$ isn't equal to $f(g_1) f(g_2)$, but that there's a homotopy relating them which itself is required to satisfy certain axioms. This homotopy is an embellished version of a $2$-cocycle.)

An earlier version of this answer claimed a counterexample for compact Lie groups (the above example with $G$ finite) that doesn't work; thanks to Omar Antolín-Camarena for pointing this out. I don't know enough about e.g. the homotopy theory of $BSU(2)$ to write down a counterexample for compact Lie groups. I might start by looking at $G = S^1, H = SU(2)$.