Yes, this follows by the de la Vallée-Poussin sufficient condition for the uniform integrability theorem. Indeed, your integral is $$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma},$$ where the $X_i$'s are independent standard normal random variables. We have $|X_i-X_j|\le2\|X\|\le1+2\|X\|$ for all $i,j$, where $X=(X_1,\dots,X_n)$ and $\|X\|=\sqrt{\sum_1^n X_i^2}$, and we have $n(n-1)/2$ factors $|X_i-X_j|^{2\gamma}$ in the product. So, $$\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}\le(1+2\|X\|)^{2\gamma \,n(n-1)/2}\le(1+2\|X\|)^C$$ for some real $C>0$ and all large enough $n$, given that $\gamma=o(1/n^2)$.
Note also that $\|X\|^2$ has the gamma distribution with parameters $n/2$ and $2$ and hence $E(1+2\|X\|)^{2C}=O(E(\|X\|^{2C})=O(n^C)$. Also, $(1+2t)^{2C}/(1+2t)^C\to\infty$ as $t\to\infty$. Also, $$\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma} =\exp\Big({2\gamma}\sum_{1\le i<j\le n}\ln|X_i-X_j|\Big)\tag{1}$$ and $$E\Big|\sum_{1\le i<j\le n}\ln|X_i-X_j|\Big|\le\sum_{1\le i<j\le n}E|\ln|X_i-X_j|\,|=O(n^2), $$ so that, by Markov's inequality, the condition $\gamma=o(1/n^2)$, and (1), $$\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma} \to1$$ in probability. Thus, by the uniform integrability, $$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}\to1.$$