One may modify Richard's attempt to get an example, using a [method of Stallings][1]. 

Consider the [Baumslag-Solitar group][2] $BS(1,2)=\langle a,b | bab^{-1}=a^2\rangle$,
with homomorphism $\phi: BS(1,2) \to H_1(BS(1,2)) = \mathbb{Z}$ given by $b\mapsto 1, a\mapsto 0$. Then $ker(\phi)=\langle b^k a b^{-k} \rangle \cong \mathbb{Z}[\frac12]$, so $ker(\phi)$ is infinitely generated. 
Take the double of $BS(1,2)$ along $ker(\phi)$, $Dub(BS(1,2),ker(\phi))$. 
This gives a group with infinite presentation 
$$Dub(BS(1,2),ker(\phi))=\langle b_1,b_2,a | b_1ab_1^{-1}=a^2=b_2ab_2^{-1}, b_1^kab_1^{-k}=b_2^k a b_2^{-k}\rangle.$$
This group has $H_2(Dub(BS(1,2),ker(\phi)),\mathbb{Z})$ infinitely generated, because of the Mayer-Vietoris sequence
$$H_2(BS(1,2))\oplus H_2(BS(1,2))\to H_2(Dub(BS(1,2),ker(\phi)) \to H_1(ker(\phi)) \to H_1(BS(1,2))\oplus H_1(BS(1,2)).$$
The homomorphism
$\mathbb{Z}[\frac12]=H_1(ker(\phi))\to H_1(BS(1,2))\oplus H_1(BS(1,2))=\mathbb{Z}^2$ is trivial, and therefore the homomorphism $H_2(Dub(BS(1,2),ker(\phi)))\to \mathbb{Z}[\frac12]$ is 
onto. Also, $H_2(BS(1,2))=0$ since it has an aspherical presentation complex with $H_2=0$.
This  implies that $H_2(Dub(BS(1,2),ker(\phi))=\mathbb{Z}[\frac12]$, and is therefore infinitely generated. 

Now, there is an automorphism of $BS(1,2)$ fixing $b\mapsto b$ and sending $a\mapsto a^2$ (corresponding to multiplication by $2$ on the subgroup $ker(\phi)\cong \mathbb{Z}[\frac12]$). This automorphism extends to $Dub(BS(1,2), ker(\phi))$. Take the mapping torus of this automorphism gives the group presentation
$$\langle b_1,b_2,a,x| b_1ab_1^{-1}=a^2=b_2ab_2^{-1}, xb_1=b_1x, xb_2=b_2x, xax^{-1}=a^2\rangle.$$
This group then is finitely presented, whose kernel of homomorphism to $\mathbb{Z}$ given by $x\mapsto 1, a, b_1, b_2\mapsto 0$ is finitely generated with infinitely generated $H_2(Dub(BS(1,2),ker(\phi)), \mathbb{Z})$. 

 

  [1]: http://www.ams.org/mathscinet-getitem?mr=158917
  [2]: http://en.wikipedia.org/wiki/Baumslag%25E2%2580%2593Solitar_group