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 $1/p$ -- so that $C(s,2)$ must depend, not on $s$, but on $p$.
$\newcommand{\si}{\sigma}\newcommand{\al}{\alpha}\newcommand\la\lambda$We have
\begin{equation*}
|\al|=h^{-1-d/2}|g(1)-g(0)|\le h^{-1-d/2}\sup_{s\in[0,1]}|g'(s)|,
\end{equation*}
where
\begin{equation*}
g(s):=\frac{G((\sqrt h\si(s))^{-1}y(s))}{\sqrt{\det a(s)}},
\end{equation*}
$y(s):=u-hb(s)$, $u:=x'-x$,
$\si(s):=(1-s)\si(t,x)+s\si(t,x')$, $b(s):=(1-s)b(t,x)+sb(t,x')$, $a(s):=(1-s)a(t,x)+sa(t,x')$.
For $s\in[0,1]$,
\begin{equation*}
\frac{g'(s)}{g(s)}=g_1+g_2+g_3,
\end{equation*}
where
\begin{equation*}
g_1:=-\frac{\det a(1)-\det a(0)}{2a(s)}, \quad
g_2:=-(b(1)-b(0))^\top a(s)^{-1}y(s),
\end{equation*}
\begin{equation*}
g_3:=\frac1h\, y(s)^\top a(s)^{-1}(a(1)-a(0))a(s)^{-1}y(s).
\end{equation*}
We have $\la^{-1}I\le a(s)\le\la I$ and hence
\begin{equation*}
\la^{-1}\le |a(s)|\le\la, \quad \det a(s)\ge\la^{-d},
\end{equation*}
and, in view of this answer, $|\det a(1)-\det a(0)|\le\la^{d-1}d\,|a(1)-a(0)|\le\la^{d-1}d\,L|u|^\eta$. So,
\begin{equation*}
|g_1|\ll|u|^\eta;
\end{equation*}
here and in what follows, $A\ll B$ and $B\gg A$ mean that $A\le CB$ for some real $C>0$ depending only on $d,\la,L,T$ -- given that $h\in(0,T]$. Next,
\begin{equation*}
|g_2|\le2L\la(|u|+hL)\ll|u|+h,
\end{equation*}
\begin{equation*}
|g_3|\le\frac1h\,\la^2 L|u|^\eta (|u|+hL)^2\ll\frac{|u|^{\eta+2}}h+|u|^\eta,
\end{equation*}
since $h\in(0,T]$.
So,
\begin{equation*}
\frac{h^{1+d/2}|\al|}{g(s)}\ll |u|^\eta+|u|+h+\frac{|u|^{\eta+2}}h.
\end{equation*}
Next,
\begin{equation*}
g(s)\le\la^{d/2}\exp\Big(-\frac{|u|^2/2-(Lh)^2}{2\la h}\Big)
\ll\exp\Big(-\frac{|u|^2}{4\la h}\Big)
\ll h^{d/2}e^{-c|u|^2/h}p_c(h,x,x')
\end{equation*}
for $c:=\frac1{6\la}$. So,
\begin{equation*}
\frac{|\al|}{h^{-1+\eta/2} p_c (h,x,x')}
\ll \Big(|u|^\eta+|u|+h+\frac{|u|^{\eta+2}}h\Big)h^{-\eta/2}e^{-c|u|^2/h}. \tag{10}\label{10}
\end{equation*}
Further,
\begin{equation*}
|u|^\eta h^{-\eta/2}e^{-c|u|^2/h}=(|u|^2/h)^{\eta/2}e^{-c|u|^2/h}
\le\max_{w\ge0}w^{\eta/2}e^{-cw}\ll1; \tag{20}\label{20}
\end{equation*}
\begin{equation*}
|u| h^{-\eta/2}e^{-c|u|^2/h}=h^{(1-\eta)/2}(|u|^2/h)^{1/2}e^{-c|u|^2/h}
\le T^{(1-\eta)/2}\max_{w\ge0}w^{1/2}e^{-cw}\ll1 \tag{30}\label{30}
\end{equation*}
provided that
\begin{equation*}
\eta\le1; \tag{$*$}\label{*}
\end{equation*}
\begin{equation*}
h\,h^{-\eta/2}e^{-c|u|^2/h}\le h^{1-\eta/2}\ll1,
\end{equation*}
again provided \eqref{*};
\begin{equation*}
\frac{|u|^{\eta+2}}h h^{-\eta/2}e^{-c|u|^2/h}=(|u|^2/h)^{\eta/2+1}e^{-c|u|^2/h}
\le\max_{w\ge0}w^{\eta/2+1}e^{-cw}\ll1. \tag{40}\label{40}
\end{equation*}
Now the desired inequality
\begin{equation*}
|\al|\ll h^{-1+\eta/2} p_c (h,x,x') \tag{50}\label{50}
\end{equation*}
follows from \eqref{10}, \eqref{20}, \eqref{30}, and \eqref{40}, provided \eqref{*}. $\quad\Box$
Without condition \eqref{*}, inequality \eqref{50} will fail to hold in this generality.
E.g., suppose that for all $t,x$ we have $\si(t,x)=I$ and $|b(t,x)|\in[0,1]$, so that $a(t,x)=I$ and your conditions 1 and 2 hold with $\la=1$, $L=2$, and any real $\eta>0$. Suppose also that $x=0$, $|x'|=\sqrt h$, $b(t,0)=0$, and $b(t,x')=-x'/|x'|$. Then
\begin{equation}
\frac{|\al|}{h^{-1+\eta/2} p_c (h,x,x')}\asymp h^{1/2-\eta/2}\to\infty
\end{equation}
as $h\downarrow0$ if $\eta>1$. $\quad\Box$
Best Answer
Let $g_n(t):=f_n (\tau^n_t)$; here and in what follows, $t\in[0,T]$. Then $$g_n(t)=f_n (\tau^n_t)\le f_n(0)+\int_0^{\tau^n_t} g_n(s)\,ds \le f_n(0)+\int_0^t g_n(s)\,ds,$$ because $\tau^n_t\le t$ and $g_n\ge0$. So, by Grönwall's inequality, $$g_n(t)\le f_n(0)e^t$$ and hence $$f_n(t)\le f_n(0)+\int_0^t g_n(s)\,ds \le f_n(0)e^t\le Cf_n(0),$$ where $C:=e^T$. $\quad\Box$