Here is an approach:
We consider the following sum in the limit $n\rightarrow\infty$ (which we denote by $S$) :
$$
S_n=\frac{1}{\binom{2n}{n}}\sum_{k=0}^n(-1)^k\binom{2n}{n+k}\frac{1}{x^2+\pi^2 k^2}
$$
We split the range of summation at some $\delta$ such that $1<<\delta<<n$ to get $S_n=S_{2,n}+S_{1,n}$ and observe that $k<<n$ in $[0,\delta]$ so we can expand the binomial around small $k/n$ which means that
$$
S_{1,n}=\sum_{k=0}^{\delta}(-1)^k\left(1+O(k^2/n^3)\right)\frac{1}{x^2+\pi^2 k^2}
$$
since $|\sum_{k=0}^{\delta}(-1)^k\frac{k^2}{x^2+\pi^2 k^2}|<\sum_{k=0}^{\delta}\frac{k^2}{x^2+\pi^2 k^2}<\sum_{k=0}^{\delta}1=\delta$ we can bound the $O(k^2/n^3)$ term as follows
$$
S_{1,n}=\sum_{k=0}^{\delta}(-1)^k\frac{1}{x^2+\pi^2 k^2}+o(\delta/n^3)
$$
Taking limits we get that $S_{1,n}$ approaches a constant to determined later
$$
S_1=\sum_{k=0}^{\infty}(-1)^k\frac{1}{x^2+\pi^2 k^2}\,\,\,\, (*)
$$
Next we show that the tail sum $S_{2,n}$ approaches zero so that the limit in question is indeed given by $(*)$. To this end, observe that on $[\delta,n]$ we have $\pi k >> x$ so we can expand the fraction around large $k$. We find
$$
S_{2,n}=\frac{1}{\binom{2n}{n}}\sum_{k=\delta}^n(-1)^k\binom{2n}{n+k}\left(\frac{1}{\pi^2 k^2}+O(1/k^4)\right)
$$
since the binomial coefficient has a maximum at $k=0$ we can bound $\frac{\binom{2n}{n+k}}{\binom{2n}{n}}\leq1$ so
$$
|S_{2,n}|<\frac{1}{\pi^2}\sum_{k=\delta}^n\frac{1}{k^2}\sim \frac{1}{\pi^2} \int_{\delta}^n\frac{dk}{k^2}\sim \frac{1}{\pi^2\delta}
$$
so, choosing $\delta$ to be a (slowly enough) increasing sequence of $n$ we have indeed that
$$
S_2=0\,\,\,\,(**)
$$
Note that the $O(1/k^4)$ are vanishing even faster and can therefore also be neglected.
Putting anything $(*)$ and $(**)$ together we get
$$
S=S_1=\frac{1}{2x^2}+\frac{1}{2\sinh(x)x}
$$
were the last equality follows from Mittag-Lefflers Theorem (or the product expansion of sine)
Best Answer
We find in H. Wilf's Generatingfunctionology formula (2.5.15) providing a generating function of the shifted central binomial coefficients in the form \begin{align*} \frac{1}{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^k=\sum_{n}\binom{2n+k}{n}x^n\tag{1} \end{align*}