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 $q>2$. 

If, finally, $q=2$, then $k=1$ and the only value of $C(s,q)$ such that \eqref{1} holds for all real $r\ge0$ is $p$ -- so that $C(s,2)$ must depend, not on $s$, but on $p$.