Yes, this follows by the [de la Vallée-Poussin sufficient condition for the uniform integrability][1]. 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 $N:=n(n-1)/2$, $X:=(X_1,\dots,X_n)$, and $\|X\|:=\sqrt{\sum_1^n X_i^2}$, and then 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, 
$$E\Big[\Big(\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}\Big)^2\Big]=O(1).$$
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, 
$$E\prod_{1\le i<j\le n}|X_i-X_j|^{2\gamma}\to1.$$


  [1]: https://en.wikipedia.org/wiki/Uniform_integrability#Relevant_theorems