The answer to the question is yes. Here are two examples, the first one answers the question as it is posted and in the second one I assume that $N(H)$ is the normal closure of $H$ in $G$. The second example seems to be more interesting, in the sense that one have to do more computation on Schreier graph, compering to the first example, where the computation of the growth of the Schreier graph of $G/H$ is straightforward.
If $N(H)$ is normalizer of $H$, then the following example works. $G:=\mathbb{Z}\times\mathbb{F}_2$ and $H$ is a subgroup $(e,H_0)$, where $H_0$ is of finite index but not normal in $\mathbb{F}_2$. Then index of $N(H)=\mathbb{Z}\times N_{\mathbb{F}_2} (H_0)$ is finite and not equal to $1$, since $N_{\mathbb{F}_2}(H_0)\neq \mathbb{F}_2$, because $H_0$ is not normal in $\mathbb{F}_2$.
If $N(H)$ is normal closure of $H$ in $G$, then the following example works. Let $\mathbb{F}_2$ be the free group on two generators $a$ and $b$. Define a homomorphism $\phi:\mathbb{F}_2\rightarrow Aut(\mathbb{Z})=\mathbb{Z}_2$ by $\phi(a)=0,\phi(b)=1$. Then the question is valid for the semidirect product $G=\mathbb{F}_2\ltimes_{\phi} \mathbb{Z}$. The Schreier graph of $G/H$ has polynomial growth and the normal closure of $H$ is $\mathbb{F}_2\ltimes_{\phi} 2\mathbb{Z}$.