Interestingly the structure of $I(2M)$ is similar to this question that OP asked before. Indeed, using functional equation of $\Gamma$ one have
$$\small I(2M)=\int_0^{\frac{1}{2}} x \sin (\pi x) \Gamma \left(M+x-\frac{1}{2}\right) \Gamma \left(M-x-\frac{1}{2}\right) \, dx=\int_0^{\frac{1}{2}} \pi x \tan (\pi x) \prod _{k=0}^{M-2} \left(\left(k+\frac{1}{2}\right)^2-x^2\right) \, dx$$
Where the primitive of RHS is evaluable through polylogarithms. The following command calculates $I(2M)$ for arbitrary $M$:
Table[Integrate[Pi x Tan[Pi x] Product[((k + 1/2)^2 - x^2), {k, 0, M - 2}], {x, 0,
1/2}], {M, 2, *Large number*}] // Expand
For a explicit calculation, one have (try it)
$$\small I(2 M)=\pi \sum _{j=0}^{M-2} \left(\sum_{1\leq a_1<\cdots<a_{M-2-j}\leq M-2} \prod_{k=1}^{M-2-j}(a_k+\frac12)^2\right) (-1)^j F_1(j)$$
Where
$$\small F_1(j)=\sum _{k=0}^{2 j+1} (-1)^k F_2(k+1) \left(\frac{1}{2}\right)^{2 j-k+2} \left(\binom{2 j+1}{k}+\binom{2 j+2}{k}\right)+F_2(2 j+3)$$
Where
$$\small F_2(k)=-\frac{k \sum _{l=0}^{k-1} (-1)^l F_3(l) \left(\frac{\pi }{2}\right)^{k-l-1} \binom{k-1}{l}}{\pi ^k}$$
Where
$$\small F_3(l)=\Re\left(\frac{(-1)^{l-1} \left(1-2^{-l-1}\right) l! \zeta (l+2)+\sum _{m=0}^l \frac{(-1)^{m-1} l! (i \pi )^{l-m} \zeta (m+2)}{(l-m)!}}{(2 i)^{l+1}}\right)$$
Enjoy (though I think a simple expression of $c_{n,d}$ in OP's (correct) speculation may not exist, other than the trivial observation that $c_{n,d}\in \mathbb Q/\pi^{2n}$)!
