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.