As stated in previous answers, the answer to the question is no.  Here is another viewpoint, which gives additional information about the relationship between $G$ and $BG$:

A pair $(G,BG)$ where $G$ is a topological group homotopy equivalent to a finite CW complex, and $BG$ its classifying space, is called a *finite loop space* in the literature. 

There has been a lot of work on these spaces, since they are generalizations of compact Lie groups. There is by now even a classification of (connected) finite loop spaces; see Section 3 of my survey paper http://www.math.ku.dk/~jg/papers/icm.pdf 

As pointed out in the previous answers, to get that $BG$ does not have a finite dimensional model, it is easy to reduce to the case where $G$ is connected (using that $BG$ is not finite for $G$ a finite group), so let's assume this. 

Now, by the structure of Hopf algebras (Milnor-Moore) $H^*(G;{\mathbb Q})$ is an exterior algebra on a number $r$ of odd dimensional generators. The number $r$ is called the (rational) *rank*, and agrees with the usual notion of rank of compact Lie groups. Hence by a spectral sequence argument $H^*(BG;{\mathbb Q})$ is a polynomial algebra on $r$ generators. So the non-finiteness of $BG$ is a corollary of the following well-known fact:

>> **Claim** A non-contractible connected finite loop space (or even H-space) has positive rank $r$.

*Proof of claim:* The claim e.g., follows from a more general statement saying the rank is also equal to the number of odd degree generators for the mod $p$ cohomology (see Kane: Homology of Hopf Spaces Section 13-3, which uses the Bockstein spectral sequence to deduce this). To get the more limited statement of the rank being positive one can give a more pedestrian argument: Suppose that $G$ is non-contractible. Then (since its a simple space and a CW complex) $\tilde H^*(G;{\mathbb Z}) \neq 0$. If this cohomology is torsion free, then the rational cohomology is non-trivial, and we are done. So suppose there is torsion. If there is non-trivial $p$-torsion in $\tilde H^*(G;{\mathbb Z})$ then, by the universal coefficient theorem,  $H^*(G;{\mathbb F_p})$ has cohomology in two consecutive degrees. So $H^*(G;{\mathbb F_p})$  has to have generators as a ring in odd degrees. But then the structure of Hopf algebras algebras shows that the Euler characteristic of $G$ has to be zero (since $H^*(G;{\mathbb F_p})$ contains an exterior tensor summand on a odd degree class, if $p$ is odd, and a truncated polynomial algebra on an odd degree class, truncated at $x^{2^k}$ for some $k$, if $p=2$). But $\chi(G) =0$ implies $\tilde H^*(G;{\mathbb Q}) \neq 0$ as wanted.