Assuming you are using the principal branch of the $s$'th root, $u(z)$ is indeed a subharmonic function. You do need to "check the branch cut", but it works out ok.

Consider a point $p$ where $f(p)$ is on the negative real axis (so $p$ is on a branch cut of $f(z)^{1/s}$), and let $v(z)$ be the version of $\Re f(z)^{1/s}$ obtained by moving the branch cut slightly away from $p$ in some direction. Then $v(z)$, being the real part of an analytic function, is harmonic near $p$, and $u(z) \ge v(z)$. So for small $r > 0$,
$$u(p) = v(p) = \frac{1}{2\pi} \int_0^{2\pi} v(p + r e^{i\theta})\ d\theta \le
\frac{1}{2\pi} \int_0^{2\pi} u(p + r e^{i\theta})\ d\theta $$