$\newcommand\ep\varepsilon$The answer is yes. Indeed, we have $L(f_n)\to L(f)$ in $L^2$ as $n\to\infty$, where $L$ denotes the Laplace transform. For real $\ep>0$, let $$f_{n,\ep}:=f_n*\psi_\ep,\quad f_{\ep}:=f*\psi_\ep,$$ where $\psi_\ep$ is the p.d.f. of the mean-zero normal distribution with variance $\ep^2$ (here we extend $f_n$ and $f$ to $\Bbb R$ by $0$). Then $$L(f_{n,\ep})=L(f_n)L(\psi_\ep)\to L(f)L(\psi_\ep)=L(f_\ep)$$ in $L^2$ as $n\to\infty$. So, inverting the Laplace transform, we get $f_{n,\ep}\to f_{\ep}$ pointwise as $n\to\infty$, which implies $C_{n,\ep}\to C_{\ep}$ pointwise and hence in $L^2$ as $n\to\infty$, where $C_{n,\ep}$ and $C_{\ep}$ are the c.d.f.'s corresponding to the p.d.f.'s $f_{n,\ep}$ and $f_{\ep}$.
Also, $C_\ep\to C$ and $C_{n,\ep}\to C_n$ in $L^2$ uniformly in $n$ as $\ep\downarrow0$. Thus, $C_n\to C$ in $L^2$.