If you just want $S'$ with the property that $C\to S'$ has connected fibers, then $S'=S$ suits perfectly as said Matthieu in the comments.
In general, let $f : C\to S$ be a proper surjective morphism of noetherian schemes. Usually, the Stein factorization refers to $S'=\mathrm{Spec}(f_*O_X)$ (it has the advantage to be canonical). This $S'$ is not always isomorphic to $S$: consider $S'\to S$ the normalization morphism of an integral curve with a cusp over $\mathbb C$ and $C=S'\times_{\mathbb C} \mathbb P^1_{\mathbb C}$. However it is true that $S'\to S$ is an isomorphism on an open subset under mild hypothesis.
(a) If $S$ is reduced and the fiber of $f$ at a generic point $\xi$ of $S$ is geometrically reduced and geometrically connected, then $S'\to S$ is an isomorphism above an open neighborhood of $\xi$.
Proof: As the question is local on $S$, we can suppose $S=\mathrm{Spec}(R)$ is affine. As $S'\to S$ is finite, $S'=\mathrm{Spec}(R')$. Because $S$ is reduced at $\xi$, $O_{S,\xi}$ is equal to the residue field $k(\xi)$ of $S$ at $\xi$. So the canonical map $R\to O_{S,\xi}=k(\xi)$ is a localization hence flat and we have
$$R'\otimes_R O_{S,\xi}= H^0(C, O_C)\otimes_R k(\xi)=H^0(C_{\xi}, O_{C_{\xi}})=k(\xi)=O_{S,\xi}.$$
So the finite morphism $S'\to S$ is an isomorphism when localized at $\xi$. By standard arguments, this isomorphism propagates above an open neighborhood of $\xi$.
(b) Suppose $f$ is flat and some geometric fiber $C_{\bar{s}}$ is connected and reduced. Then $S'\to S$ is an isomorphism above an open neighborhood of $s$.
Proof. We have $H^0(C_s, O_{C_s})=k(s)$. Then the statement is just EGA III, 7.8.8.
As a corollary of (b):
(c) If $f$ is flat with reduced and connected geometric fibers, then $S'\to S$ is an isomorphism.
$C\to S$are integral, hence connected ; thus$S'=S$does the job, doesn't it ? $\endgroup$