I am also very interested in this type of integrals. For $\gamma$ positive integer, and $\alpha_i\in\mathbb{R}$ (but perhaps the latter condition may be relaxed), I feel that Heine's theorem for (hyper)determinants might be helpful (see e.g. eq. (13) in http://arxiv.org/pdf/0912.1228.pdf). If you call your integral $I(\alpha_1,\ldots,\alpha_n)$, this is symmetric under the exchange $\alpha_i\to\alpha_j$, so it should be equal to
$$
I(\alpha_1,\ldots,\alpha_n)=\frac{1}{n!}\sum_{\sigma}\int_{-\infty}^\infty\cdots\int_{-\infty}^\infty \prod_{i=1}^n d t_i\ e^{-\frac{t_i^2}{2}+a_{\sigma(i)}t_i}\prod_{i<j}|t_i-t_j|^{2\gamma}\ ,
$$
where $\sigma$ is a permutation of the first $n$ integers. Then you can rewrite your integral introducing a *permanent*
$$
I(\alpha_1,\ldots,\alpha_n)=\frac{1}{n!}\int_{-\infty}^\infty\cdots\int_{-\infty}^\infty \prod_{i=1}^n d t_i\ e^{-\frac{t_i^2}{2}}\mathrm{perm}(e^{a_i t_j})\prod_{i<j}|t_i-t_j|^{2\gamma}\ .
$$
This should be then written as a hyperdeterminant (again, only for $\gamma$ positive integer), or as a sum over permutations of conventional determinants. For example, for $\gamma=1$ this should read
$$
I(\alpha_1,\ldots,\alpha_n)=\sum_{\sigma}\det\left(\int_{-\infty}^\infty dx\ e^{-x^2/2+a_{\sigma(i)}x}x^{i+j-2}\right),
$$
(if I am not mistaken). The integral can be computed, and perhaps some progress on the determinant evaluation can be achieved. It is admittedly not a very efficient nor complete solution, but I hope it might help.