This inequality cannot hold in general. Indeed,
\begin{equation*}
|\nabla(f^{\frac{p+q-2}{p}})|^p=k^pf^{q-2}|\nabla f|^p,
\end{equation*}
where
\begin{equation*}
k:=\frac{p+q-2}p=1+\frac{q-2}p\ge1.
\end{equation*}
So,
at all points where $g>0$, $|\nabla g|>0$, and $|\nabla f|>0$, we can rewrite the inequality in question as
\begin{equation*}
r-1\le C(s,q)(r^p-k^p) \tag{1}\label{1}
\end{equation*}
where
$r:=a/b$, $a:=f/g$, and $b:=|\nabla f|/|\nabla g|$.
In general, $r$ can take any nonnegative real value.
Letting now $r\to\infty$ in \eqref{1}, we get $C(s,q)>0$. Letting then $r=1$, we get a contradiction: $0\le C(s,q)(1-k)<0$ if $k>1$, that is, if $q>2$.