This is not a satisfactory answer, but in the absence of anything better I will put it on record. It would be nice if someone could come up with a more illuminating answer.
Computer experiments strongly indicate that the answer is yes, and I could write down a proof by hand if forced, but it would be technical and tedious, so I would prefer not to unless this is really important!
In fact there appear to be infinitely many finite index subgroups $H$ with minimal generator number $d(H) = 4$, which is best possible since $d(G/[G,G])=4$.
Indeed, wth increasing index $n$, we find subgroups $H$ of that index with all possible values of $d(H)$ between $4$ and $m(n)$, where $m(n)$ increases with $n$. For example, for $n=8$ (which is the largest index for which I can easily compute all subgroups of that index), there are subgroups $H$ of index $8$ with $d(h) = d$ for all $d$ with $4 \le d \le 18$.
It was relatively easy to find a sequence of subgroups $H$ of increasing index with $d(H)=4$. The subgroup $$\langle xy,yz,zw,x^mw^{-m} \rangle$$ has index $2m$ whereas $$\langle xy,yz,zw,x^mw^{-m+1} \rangle$$ has index $2m-1$. The first of these has right transversal $$\{ x^k,y^l : 0 \le k \le m, 1 \le l < m \}$$ and the second has right transversal $$\{ x^k,y^l : 0 \le k < m, 1 \le l < m \}.$$ As I said, I could write down a hand proof of that, but would prefer not to.
It is perhaps worth noting that for a group $\langle X \mid R \rangle$ with $|X| - |R| \ge 2$, the answer to the question would be no, because the Reidemieister-Schreier process generates a presentation of a subgroup $H$ of index $n$ with $(n-1)|X|+1$ generators, and $n|R|$ relators, so if $|X|-|R| \ge 2$ then the deficiency of the presentation would increase with increasing index, and hence so would $d(H)$.
