$\newcommand{\R}{\mathbb R} $$\newcommand{\R}{\mathbb R}$Yes, this is true for any Lipschitz compactly supported function $f$.
Indeed, we have $f(x)=0$ for some real $R>0$ and all $x\in B_R^c$, where $B_R^c:=\R^n\setminus B_R$ and $B_R$ is the closed ball of radius $R$ centered at $0$, and $|f(x)-f(y)|\le L|x-y|$ for some real $L>0$ and all $x,y$. So, theThe double integral in question is \begin{equation} I:=\int_{\R^n}dx \int_{\R^n}dy\,\frac{|f(x)-f(y)|^p}{|x-y|^{n+sp}} =2I_1+I_2, \end{equation} where \begin{equation} I_1:=\int_{B_{2R}}dx \int_{B_{2R}^c}dy\,\frac{|f(x)|^p}{|x-y|^{n+sp}}, \quad I_2:=\int_{B_{2R}}dx \int_{B_{2R}}dy\,\frac{|f(x)-f(y)|^p}{|x-y|^{n+sp}}. \end{equation} Next, \begin{equation} I_1=\int_{B_R}dx \int_{B_{2R}^c}dy\,\frac{|f(x)|^p}{|x-y|^{n+sp}} \le M^p \int_{B_R}dx \int_{B_R^c}\frac{dz}{|z|^{n+sp}}<\infty, \end{equation} where $M$ is ana finite upper bound on $|f|$, and \begin{equation} I_2\le L^p\int_{B_{2R}}dx \int_{B_{2R}}\frac{dy}{|x-y|^{n-(1-s)p}} \le L^p\int_{B_{2R}}dx \int_{B_{4R}}\frac{dz}{|z|^{n-(1-s)p}}<\infty. \end{equation} Thus, $I<\infty$. $\quad\Box$