The answer here is negative. For a surjective homomorphism $h:S(X)\to\mathbb Z$ its kernel $N$ is a normal subgroup of $S(X)$. By the Baer-Schreier-Ulam Theorem, the subgroup $N$ is equal either to the subgroup $Alt(X)$ of even finitely supported permutations or to the subgroup $S_\kappa(X)$ of permutations having support of cardinality $<\kappa$ for some infinite cardinal $\kappa\le|X|$.
Since $X$ is infinite, we can choose a family $\{x_{n,\alpha}\}_{(n,\alpha)\in\omega\times|X|}$ in $X$.
For every permutation $\pi\in S_\omega$ of $\omega$ define the permutation $\bar\pi\in S_X$ such that $\bar\pi(x_{n,\alpha})=x_{(\pi(n),\alpha)}$ for $(n,\alpha)\in\omega\times|X|$ and $\bar\pi(x)=x$ for any $x\in X\setminus\{x_{n,\alpha}:(n,\alpha)\in\omega\times|X|\}$. It is clear that $e:S_\omega\to S_X$, $e:\pi\mapsto\bar\pi$ is a group homomorphism whose image $e(S_\omega)$ in $S_X$ is disjoint with the subgroup $S_{|X|}(X)\supset N$ and hence the composition $h\circ e:S_\omega\to\mathbb Z$ is injective, which is a desirable contradiction.