You probably meant to assume $R$ is noetherian. And Hironaka seems to have suppressed his appeal to Zariski's Main Theorem, as we'll see below (or maybe someone else sees a more elementary procedure, which is certainly possible).

Consider the schematic support ${\rm{Spec}}(S/I)$ of the $S$-finite $A$ in ${\rm{Spec}}(S)$ (here, $I$ is the annihilator ideal of $A$ in $S$). Its special fiber over ${\rm{Spec}}(R)$ has underlying reduced scheme that coincides with that of the schematic support of $A/mA$ over $S/mS$, and this latter schematic support is $k$-finite (with $k := R/m$) since $A/mA$ is $k$-finite. Thus, $S/I$ has $k$-finite special fiber over $R$ since such finiteness is insensitive to killing nilpotents.

Since $R$ is *noetherian*, so $I$ is finitely generated, there is a residually trivial etale neighborhood $({\rm{Spec}}(R'),\xi)$ of $(m,z) \in {\rm{Spec}}(R[z])$ such that $I$ has a finite set of generators coming from $R'$ (via the unique map $R'_{\xi}\rightarrow S$ over $R[z]_{(m,z)}$). Letting $I'\subset R'$ be the ideal generated by those elements of $R'$, we see that $R'/I'$ has henselization $S/I$ at $\xi$ with $k$-finite special fiber over $R$. But henselization is compatible with quotients, so the special fiber of $R'_{\xi}/I'$ over $R$ is an essentially finite type local $k$-algebra with $k$-finite henselization, so $R'_{\xi}/I'$ has $k$-finite special fiber over $R$. In other words, ${\rm{Spec}}(R'/I') \rightarrow {\rm{Spec}}(R)$ is *quasi-finite* at $\xi$.

By openness of the quasi-finite locus of a finite type map between noetherian schemes, we can localize $R'$ around $\xi$ so that $R'/I'$ is a quasi-finite $R$-algebra. Hence, by Zariski's Main Theorem, ${\rm{Spec}}(R'/I')$ is Zariski-open in a *finite* $R$-algebra. But $R$ is henselian, so by Hensel's Lemma (in the EGA form) applied to lifting idempotents from the special fiber we know that every finite $R$-algebra is a direct product of finite *local* $R$-algebras. Consequently, by shrinking some more around the point $\xi$ in the special fiber we can arrange that ${\rm{Spec}}(R'/I')$ is Zariski-open in a *finite local* $R$-scheme with closed point $\xi$, so this Zariski-open locus is the entire space. In other words, we get to the case that $R'/I'$ is *$R$-finite* and *local*. Thus, it is equal to its own henselization, which is $S/I$ by design.

We have proved that $S/I$ is $R$-finite, yet $A$ is an $S/I$-module, so $A$ s $R$-finite. QED