$\newcommand{\Om}{\Omega} $The answer is yes, and it is even true that in your conditions
\begin{equation*}
	I:=\int_{\Omega} \frac{f(x)^2}{(x_1-x_2)^2} \ dx_1 \ dx_2 \le C \|\nabla f \|^2_{L^2}\tag{1}\label{1}
\end{equation*}

Indeed, using the substitution $x_1-x_2=u$ and $x_1+x_2=v$ and doing some rescaling, reduce \eqref{1} to  
\begin{equation*}
	\int_0^1 \frac{g(u)^2}{u^2}\,du\le C_1 \int_0^1 g'(u)^2\,du \tag{2}\label{2}
\end{equation*}
for some universal real constant $C_1>0$ and all smooth $g$ with $g(0)=0$. But \eqref{2} is a special case of [Hardy's inequality][1], with $p=2$, so that \eqref{2} holds with $C_1=4$.

---

**Details on the reduction of \eqref{1} to\eqref{2}:** By rescaling, without loss of generality $L=1/2$. Let $F(u,v):=f(\frac{u+v}2,\frac{u-v}2)$ for $(u,v)\in U(\Om)\subseteq[-1,1]^2=S$, where $U(x_1,x_2):=(x_1-x_2,x_1+x_2)$, so that $f(x)=f(x_1,x_2)=F(x_1-x_2,x_1+x_2)$ for $x\in\Om$. Extend $F$ by $0$ to the function on $S$, which will be denoted still by $F$. Then 
\begin{equation*}
	2I=\int_{-1}^1 dv\,\int_{-1}^1 du\, \frac{F(u,v)^2}{u^2}
	=I_+ +I_-,
\end{equation*}
where 
\begin{equation*}
	I_{\pm}:=\int_{-1}^1 dv\,\int_{-1}^1 du\, \frac{F(u,v)^2}{u^2}\,1(\pm u>0). 
\end{equation*}
Letting $F_u$ denote the partial derivative of $F$ wrt $u$, using \eqref{2} with $C_1=4$, and noting that $F_u(u,v)^2\le|\nabla f(\frac{u+v}2,\frac{u-v}2)|^2$, we get 
\begin{equation*}
	I_+\le4\int_{-1}^1 dv\,\int_0^1 du\, F_u(u,v)^2 \\ 
	\le4\int_{-1}^1 dv\,\int_0^1 du\, |\nabla f(\tfrac{u+v}2,\tfrac{u-v}2)|^2. 
\end{equation*} 
Similarly, 
\begin{equation*}
	I_- 
	\le4\int_{-1}^1 dv\,\int_{-1}^0 du\, |\nabla f(\tfrac{u+v}2,\tfrac{u-v}2)|^2,  
\end{equation*} 
so that 
\begin{equation*}
	I\le2\int_{-1}^1 dv\,\int_{-1}^1 du\, |\nabla f(\tfrac{u+v}2,\tfrac{u-v}2)|^2 \\ 
=4\int_{\Om}|\nabla f|^2
=4\|\nabla f \|^2_{L^2}. 
\end{equation*} 
So, \eqref{1} holds with $C=4$ (for $L=1/2$) $\quad\Box$. 

  [1]: https://en.wikipedia.org/wiki/Hardy%27s_inequality