The answer is no: actually by Baumslag-Roseblade (1984: [BR]) either $G$ is commensurable to a product of free groups (hence has linear or quadratic Dehn function), or is not finitely presented (so the Dehn function is infinite, or not defined, as you wish). The latter case occurs if $H$ is an infinite word hyperbolic group and $f:F_m\to H$ is a non-bijective surjection and $H$ is the fibre product $$\{(g,h)\in F_m\times F_m:f(g)=f(h)\}.$$

**Edit 1** *(reply to Lee Mosher's comment, copy from a comment dated Sep 14 '12)* In general the Dehn function of $H$ is equivalent to the distortion of $G$. This is not completely formal (there's a little Van Kampen diagram cuisine) but in the case of $H=\mathbf{Z}$ it's simple to verify by hand. I saw the general result written somewhere but I can't remember right now.

**Edit 2** *(Extracted from Robert Young's comment to his reply)* There's a proof for [this non-distortion fact] mentioned in Olshanskii, Sapir [0S, Theorem 2].

References:

[BR] G. Baumslag, J. Roseblade. Subgroups of direct products of free groups. J. London Math. Soc. (2) 30 (1984), no. 1, 44-52. Journal link (restricted access). MR link

[OS] A. Ol'shanskii, M. Sapir. Length and area functions on groups and quasi-isometric Higman embeddings. ArXiv Link. Internat. J. Algebra Comput. 11 (2001), no. 2, 137-170. Journal link (restricted access). MR link