$\newcommand{\Om}{\Omega}$The answer is no. 

Indeed, by rescaling, without loss of generality $L=1$. Suppose that 
\begin{equation}
f(x_1,x_2)=g_a(|x_1-x_2|)	
\end{equation}
for $(x_1,x_2)\in\Om$, 
where $g\in W^{1,2}((-1,1))$ such that 
$g_a(u):=|u|^a$ for a real $a>0$ and all $u\in(-1/2,1/2)$. 

So, if your were true, then we would have 
\begin{equation*}
	L(a):=\int_{(-1,1)^2} \frac{g_a(u)^2}{|u|^2}\,du \\ 
	\le C_1 \int_{(-1,1)^2} (g_a(u)^2+|\nabla g_a(u)|^2)\,du=:R(a)  \tag{1}\label{1}
\end{equation*}	
for some real $C_1>0$ and all real $a>0$. 

However, for $a\downarrow0$ we have $L(a)\asymp1/a$ whereas $R(a)\asymp1$, and hence \eqref{1} fails to hold.