Going out on a limb (I may well be messing up badly), I think the answer is yes at least for the CW structure version of the question.
Proof:
Choose a finite presentation of $F$ with generating set $G$ and relation set $R$, and consider the induced $\pi_1$-isomorphism $X = (S^1)^{\vee G} \cup_{(S^1)^{\vee R}} \ast \to BF$. Note that $X$ has finitely many cells. I believe that the rank over $\mathbb Z[F]$ of $H_\ast(\tilde X ; F)$$H_\ast(\tilde X ; \underline{\mathbb Z[F]})$ grows polynomially in $\ast$ (if it doesn't, then we should have a homology obstruction, giving a negative answer to the question), and. Then we should be able to mimic the construction showing that one canthe homology bound on number of cells is realized for a simply-connected space. That is, we attach free $F$-equivariant cells (i.e. cells of the form $\vee^{F} D^n_+$, with $n \geq 2$) to $\tilde X$ one-by-one, starting with $X$, building up a space $X'$ whose universal cover$\tilde X'$ which has the same $\mathbb Z[F]$$\underline{\mathbb Z[F]}$-homology as $BF$$\tilde{BF}$ and conclude by homology Whitehead (with coefficients) that $X' \to BF$ is a weak homotopy equivalence, where $X' = \tilde X'_{hF}$ has one cell for each cell of $X$ plus a cell for each $F$-equivariant cell we attached.
(I'm not 100% sure though -- in the simply-connected case, we use the Hurewicz theorem to be sure that we can always map to a homology class with a sphere... perhaps this breaks down if the relevant $F$-equivariant Hurewicz theorem fails?)