$\newcommand{\de}{\delta}\newcommand{\ep}{\varepsilon}$Yes, this is true. Moreover, the continuity and the strictness of the decrease of the $f_n$'s are not needed.

Indeed, for any nonincreasing function $g\colon[0,\infty)\to(0,1]$ and any $t\in(0,1)$, let
\begin{equation*}
g^{-1}(t):=\sup\{x\in[0,\infty)\colon g(x)\ge t\}, \tag{0}\label{0}
\end{equation*}
with $g^{-1}(t):=0$ if $\{x\in[0,\infty)\colon g(x)\ge t\}=\emptyset$. Of course, in the particular case when the function $g$ is continuously and strictly decreasing from $1$ to $0$ on $[0,\infty)$, the generalized inverse $g^{-1}$ defined by formula \eqref{1} coincides with the usual inverse of $g$.

Let us show that
\begin{equation*}
f_n^{-1}\to f^{-1} \tag{1}\label{1}
\end{equation*}
pointwise on the set
\begin{equation*}
C_*:=C\cup(0,t_\infty),
\end{equation*}
where $t_\infty:=\lim_{x\to\infty}f(x)\in[0,1]$ and $C$ is the set of all points $t\in(0,1)$ such that $f^{-1}$ is finite and continuous at $t$. Note that $f^{-1}=\infty$ on $(0,t_\infty)$.

Since the function $f^{-1}$ is nonincreasing, the set $(0,1)\setminus C_*$ is at most countable. So, it will follow that the convergence \eqref{1} is almost everywhere (a.e.), as desired.

To prove \eqref{1}, take first any $t_0\in C$ and let $x_0:=f^{-1}(t_0)$, so that $x_0\in[0,\infty)$. Take any real $\ep>0$. The condition $t_0\in C$ implies that for some real $\de=\de_\ep>0$ we have
\begin{equation*}
|f^{-1}(t_0\pm\de)-f^{-1}(t_0)|<\ep.
\end{equation*}
By the pointwise convergence $f_n\to f$, there is some natural $n_\ep=n_{\ep,\de_\ep}$ such that
\begin{equation*}
n\ge n_\ep\implies |f_n(x_0\pm\ep)-f(x_0\pm\ep)|<\de.
\end{equation*}

So, for $n\ge n_\ep$ we have the following implications:
\begin{equation*}
\begin{aligned}
&f_n^{-1}(t_0)>x_0+\ep\implies f_n(x_0+\ep)\ge t_0
\implies f(x_0+\ep)>t_0-\de \\
& \implies x_0+\ep\le f^{-1}(t_0-\de)<f^{-1}(t_0)+\ep=x_0+\ep,
\end{aligned}
\end{equation*}
a contradiction. So, for $n\ge n_\ep$,

\begin{equation*}
f_n^{-1}(t_0)\le x_0+\ep=f^{-1}(t_0)+\ep.
\end{equation*}
Similarly, for $n\ge n_\ep$ we have
\begin{equation*}
\begin{aligned}
& f_n^{-1}(t_0)<x_0-\ep\implies f_n(x_0-\ep)<t_0
\implies f(x_0-\ep)<t_0+\de \\
& \implies x_0-\ep\ge f^{-1}(t_0+\de)>f^{-1}(t_0)-\ep=x_0-\ep,
\end{aligned}
\end{equation*}
a contradiction. So, for $n\ge n_\ep$,

\begin{equation*}
f_n^{-1}(t_0)\ge x_0-\ep=f^{-1}(t_0)-\ep.
\end{equation*}
Thus, convergence \eqref{1} holds on $C$.

Take now any $t_0\in(0,t_\infty)$, so that $f^{-1}(t_0)=\infty$. Take any real $x\ge0$. Then $f(x)>t_0$. So, there is some natural $n_x$ such that for all $n\ge n_x$ we have $f_n(x)>t_0$ and hence $f_n^{-1}(t_0)\ge x$. It follows that $f_n^{-1}(t_0)\to\infty=f^{-1}(t_0)$. So, convergence \eqref{1} holds on $(0,t_\infty)$ as well, and thus it does hold on $C_*$, as claimed. $\quad\Box$