This is an older problem so I will give a fairly complete solution.
We use the physicist's convention for the Hermite polynomials.
There are many ways to show the two representations are the same.
Let's expand the integrals in small $x$ and show that they have the same series.
We start from the contour integral,
$$\begin{eqnarray*}
H_n(x) &=& \frac{n!}{2\pi i} \oint_\gamma d t\, \frac{e^{2tx -t^2}}{t^{n+1}} \\
&=& \frac{n!}{2\pi i} \sum_{k=0}^\infty \frac{(2x)^k}{k!}
\sum_{l=0}^\infty \frac{(-)^l}{l!} \oint_\gamma d t\, t^{2l+k-n-1}.
\end{eqnarray*}$$
But the contour integral is zero unless $2l+k-n = 0$, otherwise it is $2\pi i$.
Notice that $n-k$ must be even.
Thus,
$$H_n(x) = n! \sum_{k=0 \atop n-k\ \mathrm{even}}^\infty (-)^{\frac{n-k}{2}} \frac{(2x)^k}{k!(\frac{n-k}{2})!}.$$
The sum will terminate at $k=n$ since the gamma function has simple poles at $0,-1,-2,\ldots$.
Letting $l = (n-k)/2$ we find
$$\begin{equation}
H_n(x) = n! \sum_{l=0}^{\lfloor \frac{n}{2}\rfloor} (-)^l \frac{(2x)^{n-2l}}{l!(n-2l)!},\tag{1}
\end{equation}$$
the standard series representation of the Hermite polynomials.
Now for the other integral representation.
Expand the binomial series,
$$\begin{eqnarray*}
H_n(x) &=& \frac{2^n}{\sqrt{\pi}} \int_{-\infty}^\infty dt\, e^{-t^2}(x+it)^n \\
&=& \frac{2^n}{\sqrt{\pi}} \sum_{k=0}^n {n\choose k} x^k i^{n-k}
\int_{-\infty}^\infty dt\, e^{-t^2} t^{n-k}.
\end{eqnarray*}$$
The integral is zero unless $n-k$ is even.
Therefore,
$$\begin{eqnarray*}
\int_{-\infty}^\infty dt\, e^{-t^2} t^{n-k}
&=& 2\int_{0}^\infty dt\, e^{-t^2} t^{n-k} \\
&=& \Gamma\left(\frac{n-k}{2}+\frac{1}{2}\right) \\
&=& \sqrt{\pi} \frac{(n-k)!}{2^{n-k}(\frac{n-k}{2})!}.
\end{eqnarray*}$$
(Change variables, let $z = t^2$, to see the connection to the gamma function.)
Substitute this into our series for $H_n(x)$.
After the dust settles, we let $l = (n-k)/2$ and arrive back at equation (1) above.
Finally I found how to do it. I post it, if someone is interested.
\begin{align}
D_{ln}(\varkappa) &= \frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty}H_n(\tilde{x})e^{-\tilde{x}^2+\varkappa \tilde{x}}H_l(\tilde{x})\;\mathrm{d}\tilde{x} \notag\\
&= \frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty}H_n(\tilde{x})e^{-\tilde{x}^2+\varkappa \tilde{x}-\varkappa^2/4}e^{\varkappa^2/4}H_l(\tilde{x})\;\mathrm{d}\tilde{x} \notag\\
&= \frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}e^{\varkappa^2/4}\underbrace{\int_{-\infty}^{+\infty}H_n(\tilde{x})e^{-(\tilde{x}-\varkappa/2)^2}H_l(\tilde{x})\;\mathrm{d}\tilde{x}}_I
\end{align}
If we pose $x = \tilde{x}-\frac{\varkappa}{2}$ in this expression, the integral $I$ becomes
\begin{equation*}
I = \int_{-\infty}^{+\infty}H_n(x+\varkappa/2)e^{-x^2}H_l(x+\varkappa/2)\;\mathrm{d}x
\end{equation*}
We know that
\begin{equation*}
H_n(x+a) = \sum_{p=0}^n \frac{n!}{(n-p)!p!}(2a)^{n-p}H_p(x)
\end{equation*}
Hence, the integral $I$ becomes
\begin{align*}
I &= \int_{-\infty}^{+\infty} \sum_{p=0}^n \frac{n!}{(n-p)!p!}\varkappa^{n-p}H_p(x) e^{-x^2} \sum_{q=0}^l \frac{l!}{(l-q)!q!}\varkappa^{l-q}H_q(x)\;\mathrm{d}x \\
&= \sum_{p=0}^n\sum_{q=0}^l \frac{n!}{(n-p)!p!}\varkappa^{n-p}\frac{l!}{(l-q)!q!}\varkappa^{l-q}\int_{-\infty}^{+\infty}H_p(x) e^{-x^2}H_q(x)\;\mathrm{d}x \\
\end{align*}
The Hermite polynomials are orthogonal in the range $(-\infty,\infty)$ with respect to the weighting function $e^{-x^2}$ and satisfy
\begin{alignat*}{2}
&&&\int_{-\infty}^{+\infty}H_p(x) e^{-x^2}H_q(x)\;\mathrm{d}x = \sqrt{\pi}2^pp!\;\delta_{pq} \\
&\Rightarrow\quad&& I = \sum_{p=0}^n\sum_{q=0}^l \frac{n!}{(n-p)!p!}\frac{l!}{(l-q)!q!}\varkappa^{n+l-p-q}\cdot \sqrt{\pi}2^pp!\;\delta_{pq}
\end{alignat*}
As this integral is nil if we have not $p=q$, we can replace the two sums by only one that goes from 0 to $\min(n,l)$. Let us say that $n<l$. Hence, the full expression for the $D$-matrix is
\begin{align}
D_{ln}(\varkappa) &= \frac{1}{\sqrt{2^nn!}\sqrt{2^ll!}}\frac{1}{\sqrt{\pi}}e^{\varkappa^2/4} \sum_{p=0}^n \frac{n!}{(n-p)!p!}\frac{l!}{(l-p)!p!}2^pp!\sqrt{\pi}\;\varkappa^{n+l-2p} \notag\\
&= \frac{\varkappa^{n+l}}{\sqrt{2^nn!}\sqrt{2^ll!}}e^{\varkappa^2/4} \sum_{p=0}^n \frac{n!}{(n-p)!p!}\frac{l!}{(l-p)!}2^p\;\varkappa^{-2p} \notag\\
&= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{n+l}e^{\varkappa^2/4} \sum_{p=0}^n \frac{l!}{(n-p)!(l-p)!p!}\left(\frac{\varkappa^2}{2}\right)^{-p} \notag\\
&= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4} \sum_{p=0}^n \frac{l!}{(n-p)!(l-p)!p!}\left(\frac{\varkappa^2}{2}\right)^{n-p}
\end{align}
Associated Laguerre polynomials $L_a^b(x)$ are given by
\begin{equation*}
L_a^b(x) = \sum_{k=0}^{a}(-1)^k \frac{(a+b)!}{(a-k)!(b+k)!k!}x^k
\end{equation*}
It suggests us to transform the expression of the $D$-matrix by posing $k=n-p$. Hence, we have
\begin{align}
D_{ln}(\varkappa) &= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4} \sum_{k=n}^0 \frac{(l)!}{(n-(n-k))!(l-(n-k))!(n-k)!}\left(\frac{\varkappa^2}{2}\right)^{n-(n-k)} \notag\\
&= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4} \sum_{k=0}^n \frac{l!}{k!(l-n+k)!(n-k)!}\left(\frac{\varkappa^2}{2}\right)^{k} \notag\\
&= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4} \sum_{k=0}^n (-1)^k\frac{([l-n]+n)!}{(n-k)!([l-n]+k)!k!}\left(-\frac{\varkappa^2}{2}\right)^{k} \notag\\
&= \sqrt{\frac{n!}{l!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{l-n}e^{\varkappa^2/4}L_n^{l-n}\left(-\frac{\varkappa^2}{2}\right)
\end{align}
It should be remembered that we had supposed that $n<l$. But that could be otherwise. In order to be general, $n_<$ and $n_>$ will be defined as $n_<=\min{(n,l)}$ and $n_>=\max{(n,l)}$ and $l-n=|l-n|$. We then have
\begin{equation}
D_{ln}(\varkappa) = \sqrt{\frac{n_<!}{n_>!}}\left(\frac{\varkappa}{\sqrt{2}}\right)^{|l-n|}L_{n_<}^{|l-n|}\left(-\frac{\varkappa^2}{2}\right)e^{\varkappa^2/4}
\end{equation}
Best Answer
With $g(x,t)$ not having any finite poles (entire in $t$) the Cauchy Coefficient Formula certainly applies. We get
$$H_n(x) = \frac{n!}{2\pi i} \int_{|t|=\epsilon} \frac{1}{t^{n+1}} \exp(-t^2+2tx) \; dt.$$
This is
$$ \frac{n!\times \exp(x^2)}{2\pi i} \int_{|t|=\epsilon} \frac{1}{t^{n+1}} \exp(-t^2+2tx-x^2) \; dt \\ = \frac{n!\times \exp(x^2)}{2\pi i} \int_{|t|=\epsilon} \frac{1}{t^{n+1}} \exp(-(t-x)^2) \; dt.$$
With the substitution $z=t-x$ we have $dz = dt$ and the radius of the small circle is preserved, and we make one turn. We find
$$\frac{n!\times \exp(x^2)}{2\pi i} \int_{|z+x|=\epsilon} \frac{1}{(z+x)^{n+1}} \exp(-z^2) \; dz$$
as claimed. The references for the CCF are Analytic Combinatorics by Flajolet and Sedgewick, e.g. pages 246 and 732.