It's only true when $l = 0$ or $q$ is odd (since both sides are $0$), otherwise it is not true.
(I'll just do the case when $l$ is even, the case when $l$ is odd is pretty much identical.)
Since $\binom{a}{b}$ is at a maximum when $b = a/2$ (or $b = (a\pm 1)/2$ if $a$ is odd) and all the terms in the summation are non-negative ($q$ is even), we have
$0 \le \sum_{k=0}^l \frac{\binom{2n-l}{n-k}}{\binom{2n}{n}} (2k-l)^q\binom{l}{k} \le \frac{\binom{2n-l}{n-l/2}}{\binom{2n}{n}} \sum_{k=0}^l (2k-l)^q\binom{l}{k}$
$\binom{a}{a/2}$ is increasing, so it is maximised by taking $a$ as large as possible, that is, choosing $l$ as small as possible; so $l = 2$ ($0$ was eliminated above). Thus \begin{align*}\frac{\binom{2n-l}{n-l/2}}{\binom{2n}{n}} \sum_{k=0}^l (2k-l)^q\binom{l}{k} &\le \frac{\binom{2n-2}{n-1}}{\binom{2n}{n}} \sum_{k=0}^l (2k-l)^q\binom{l}{k} \\ &= \frac{n}{2n(2n-1)}\sum_{k=0}^l (2k-l)^q\binom{l}{k} \\ &= \frac{1}{2(2n-1)}\sum_{k=0}^l (2k-l)^q\binom{l}{k} \end{align*}
Thus, the approximation can't hold, since the left-hand side goes to $0$ as $n \to \infty$ while the right hand side doesn't.