The pattern continues indeed except the RHS becomes more and more involved as you increase $r$. At any rate, here is what we get for the case $r=4$: $$\sum_{k=0}^n\binom{n}k\binom{n}{k+a}\binom{2k-n+a}4= \frac{(a^2-n^2-6n+11)\,(a-n)\,(a+n)}{8\,(2n-1)\,(2n-3)}\cdot \binom{2n}{n+a}.$$ Another sampler for $r=5$: $$\sum_{k=0}^n\binom{n}k\binom{n}{k+a}\binom{2k-n+a}5= -\frac{(a^2-n^2-2n+5)\,(a-n)\,(a+n)}{4\,(2n-1)\,(2n-3)}\cdot \binom{2n}{n+a}.$$ These kinds of problems can be proved by the automated tools of Zeilberger's algorithm.

The pattern continues indeed except the RHS becomes more and more involved as you increase $r$. At any rate, here is what we get for the case $r=4$: $$\sum_{k=0}^n\binom{n}k\binom{n}{k+a}\binom{2k-n+a}4= \frac{(a^2-n^2-6n+11)\,(a-n)\,(a+n)}{8\,(2n-1)\,(2n-3)}\cdot \binom{2n}{n+a}.$$ Another sampler for $r=5$: $$\sum_{k=0}^n\binom{n}k\binom{n}{k+a}\binom{2k-n+a}5= -\frac{(a^2-n^2-2n+5)\,(a-n)\,(a+n)}{4\,(2n-1)\,(2n-3)}\cdot \binom{2n}{n+a}.$$ These kinds of problems can be proved by the automated tools of Zeilberger's algorithm which relies on Wilf-Zeilberger pairs. If you are able to use the Maple symbolic software then download the algorithm to make use of it. These days, this package is also part of the routines in Maple as well as Mathematica.

The pattern continues indeed except the RHS becomes more and more involved as you increase $r$. At any rate, here is what we get for the case $r=4$: $$\sum_{k=0}^n\binom{n}k\binom{n}{k+a}\binom{2k-n+a}4= \frac{(a^2-n^2-6n+11)\,(a-n)\,(a+n)}{8\,(2n-1)\,(2n-3)}\cdot \binom{2n}{n+a}.$$ This can be proved by the automated tools of Zeilberger's algorithm.

The pattern continues indeed except the RHS becomes more and more involved as you increase $r$. At any rate, here is what we get for the case $r=4$: $$\sum_{k=0}^n\binom{n}k\binom{n}{k+a}\binom{2k-n+a}4= \frac{(a^2-n^2-6n+11)\,(a-n)\,(a+n)}{8\,(2n-1)\,(2n-3)}\cdot \binom{2n}{n+a}.$$ Another sampler for $r=5$: $$\sum_{k=0}^n\binom{n}k\binom{n}{k+a}\binom{2k-n+a}5= -\frac{(a^2-n^2-2n+5)\,(a-n)\,(a+n)}{4\,(2n-1)\,(2n-3)}\cdot \binom{2n}{n+a}.$$ These kinds of problems can be proved by the automated tools of Zeilberger's algorithm.