(Too long for a comment.) After staring hard at my question and recalling an old MSE post of mine, I made an inspired guess and found,
$$\frac{1}{\pi}=\frac{192\sqrt{2}}{(396^2+4\color{blue}\alpha)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k \color{blue}\alpha^{k-2j} \tbinom{k}{2j}\tbinom{2j}{j}\tbinom{2j}{j} \frac{58\cdot15015k+72798-37\color{blue}\alpha/4}{(396^2+4\color{blue}\alpha)^k}$$
for general $\alpha$, so turns out it is unnecessary to restrict it to powers of $2$. (Thus, $s_5(k)$ and $s_6(k)$ do have a closed-form and are integer sequences.)
Similarly, the Chudnovsky formula (a level 1 Ramanujan-Sato) generalizes to a level 9,
$$\frac{1}{\pi}=\frac{12}{(640320-4\color{blue}\alpha)^{3/2}}\sum_{k=0}^\infty\tbinom{2k}{k}\sum_{j=0}^k \color{blue}\alpha^{k-3j} \tbinom{k}{3j}\tbinom{2j}{j}\tbinom{3j}{j} \frac{163A\,k+B+1448\color{blue}\alpha/3}{(-640320+4\color{blue}\alpha)^k}$$
where $A=1114806, B=13591409$.
However, I do not have a rigorous proof for these two families and the relevant literature do not seem to address general $\alpha$.