Is there an example of two non isomorphic finite groups $G$ and $H$ such that $BG^{+}$ is homotopy equivalent to $BH^{+}$ ?
Edit: I thought I had an example, but it does not quite work. It answers a different question. I will edit this post a bit, make it Community Wiki, and leave it up just in case it is of some interest, or might help to find a complete answer.
For some finite groups $G$, $BG^+$ is equivalent to the product of the $p$completions $BG\hat{_p}$. For example, it is true if $G$ is perfect. More generally, it is true if the final term of the lower central series is a perfect subgroup, and the plus construction is taken with respect to this subgroup. So we may consider the related question whether nonisomorphic finite groups can have equivalent $p$completed classifying spaces for all primes $p$ (in the same spirit as Matthias Wendt's answer).
By a theorem of Oliver, the $p$completion of $BG$ is determined by the $p$fusion system of $G$, where the $p$fusion system of $G$ is the category whose objects are subgroups of a $p$Sylow subgroup $P$, and where morphisms are all homomorphisms that are induced by conjugation by an element of $G$.
So this question is equivalent to whether there can be nonisomorphic finite groups whose $p$fusion systems are equivalent for all primes $p$. Such examples are known to exist. I found the following one in a paper of Martino and Priddy (who attribute it to Minami): $Q_{4p}\times {\mathbb Z}/2$ and $D_{2p}\times {\mathbb Z}/4$. Here $Q_{4p}$ is the generalised Quaternion group of order $4p$, where $p$ is an odd prime. The point is that they have isomorphic $p$Sylow subgroups at all primes, namely ${\mathbb Z}/2\times {\mathbb Z}/4$ and ${\mathbb Z}/p$, and moreover the $p$fusion systems are equivalent: you get the trivial structure at the prime $2$, and the only nontrivial morphism at $p$ is the inverse homomorphism.
However, these groups are solvable but not nilpotent. So they do not have nontrivial perfect subgroups, and their lower central series do not terminate at the trivial group. So I don't believe their plus constructions are equivalent. Indeed, the plus construction does nothing to these groups. An example where the groups have the property that the lower central series terminate at a perfect subgroup would answer the original question.

3$\begingroup$ Dear Greg, here is a brief comment. Suppose $P < G$ is perfect and $Q < H$ is also perfect, and there is an equivalence between plusconstructions $BG^+ \to BH^+$. Then on $\pi_1$ it gives an isomorphism $G/P \to H/Q$, and so we get an equivalence of universal covers. I believe that the map on universal covers is equivalent to $BP^+ \to BQ^+$. Thus we might as well start by looking for examples where $G$ and $H$ are perfect already and you can use the $p$completions you describe. $\endgroup$ – Tyler Lawson Oct 9 '17 at 23:37

1$\begingroup$ Can't we create perfection with the wreath product? $S_5\ltimes G^5$ has abelianization $S_5^{ab}\times G^{ab}$ and perfect derived subgroup. And presumably this is functorial in the fusion system. $\endgroup$ – Ben Wieland Oct 10 '17 at 21:34

$\begingroup$ @BenWieland Sounds good to me. Why don't you make it into an answer :) $\endgroup$ – Gregory Arone Oct 11 '17 at 5:06
Not an answer, but here are some partial statements which at least tell us that we need to look for quite elaborate examples.
An equivalence of plus constructions implies in particular that there is an isomorphism of cohomology rings ${\rm H}^\bullet(G,\mathbb{Z}/p)\cong {\rm H}^\bullet(H,\mathbb{Z}/p)$ for all primes $p$. Then we can use the results from Quillen's papers on the spectrum of equivariant cohomology rings to relate the structure of $G$ and $H$:
 D.G. Quillen. The spectrum of an equivariant cohomology ring, I & II. Ann. Math. 94, No. 3 (1971), pp. 549–572, pp. 573–602.
For example, Corollary 2.4 of part I says that a homomorphism $\phi:G\to H$ induces a finite map on cohomology rings if and only if the kernel is a finite group of order prime to $p$. A direct consequence of knowing this for all primes $p$: if $BG^+$ is contractible, then $G$ is trivial. More generally, by a theorem of Mislin, if there is a group homomorphism inducing the plusconstruction equivalence, then this is an isomorphism.
 G. Mislin. Lannes' $T$functor and the cohomology of $BG$. Compositio Math. 86 (1993), no. 2, 177–187.
There are much more elaborate results that can be formulated for the spectrum of the equivariant cohomology ring. The Krulldimension is the rank of a maximal elementary abelian $p$group, so these would have to agree for all $p$ if we have a cohomology isomorphism. Actually, the structure of the spectrum of equivariant cohomology encodes all sorts of information about conjugacy classes of elementary abelian $p$subgroups and their Weyl groups, and all these bis of information have to agree for two groups having isomorphic group cohomology. The question remains how much information about elementary abelian $p$subgroups is needed to get an isomorphism of groups. I don't know enough about support varieties, Lannes Tfunctor or group theory to answer that, but maybe somebody else can take it from here...

3$\begingroup$ Along the same lines: the Morava $E$theory of $BG$ is the same as for $BG^+$, so we can use HKR generalized character theory to see that an equivalence $BG^+\to BH^+$ gives a bijection $[\mathbb{Z}_p^n,G]\to[\mathbb{Z}_p^n,H]$ for all $p$ and $n$. (Here $[A,G]$ is the set of conjugacy classes of homomorphisms $A\to G$, and $\mathbb{Z}_p$ is the additive group of $p$adic integers.) $\endgroup$ – Neil Strickland Oct 12 '17 at 0:40