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. Introducing $X:=(X_1,\dots,X_n)$ and $\|X\|:=\sqrt{\sum_1^n X_i^2}$, and using the arithmetic-geometric-mean inequality, we have $$\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma} \le\Big(\frac1N\,\sum_{1\le i<j\le n}|X_i-X_j|^2\Big)^{N\gamma} \\ =O\Big(\frac{\|X\|^2}n\Big)^{N\gamma}=O\Big(1+\frac{\|X\|^2}n\Big)^C $$ for any 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 $E\|X\|^{2C}=O(n^C)$. So, $$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}=O(1)$$ and, similarly, $$E\Big(\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}\Big)^2=O(1).$$ Also, obviously, $t^2/t\to\infty$ as $t\to\infty$. So, we have the uniform integrability.
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, $\sum_{1\le i<j\le n}\ln|X_i-X_j|=O(n^2)$ in probability. So, by 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, $$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}\to1.$$