After some work something has come. The answer to the question is yes, but the sums are not obvius and with not beatiful forms.
We have:
$$\frac{8K(k)E(k)}{\pi^2}=\frac{2-k^2}{\sqrt{1-k^2}}\sum_{n=0}^{\infty}\frac{(2n+1)(2n)!^3}{2^{8n}n!^6}{\left(\frac{k^4}{
k^2-1}\right)}^n,\tag{1}$$
with $-\frac{1}{\sqrt{2}}\leq k\leq\frac{1}{\sqrt{2}}.$ Where:
$$K(k)=\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2{(x)}}},\tag{2}$$
and
$$E(k)=\int_{0}^{\pi/2}\sqrt{1-k^2\sin^2{(x)}}dx,\tag{3}$$
are the complete elliptic of the first and second kind respectively.
As suggested by Bob Dobbs and Paramanand Singh we provide a proof of $(1)$.
Since we have:
$$K(k)(1-k^2)=E(k)-K'(k)(k-k^3),\tag{4}$$
where $K'(k)$ here is the derivative of $K(k)$.
Multiplying both sides by $K(k)$:
$$K^2(k)(1-k^2)+K(k)K'(k)(k-k^3)=K(k)E(k),\tag{5}$$
We recognize $K(k)K'(k)$ as $\left(\frac{K^2(k)}{2}\right)'$. Then:
$$K^2(k)(1-k^2)+\left(\frac{K^2(k)}{2}\right)'(k-k^3)=K(k)E(k).\tag{6}$$
Now we invoke Landen's transformation:
$$K(k)(1+k)=K(\frac{2\sqrt{k}}{1+k}),\tag{7}$$
and Clausen's formula (1828) (https://arxiv.org/pdf/1302.5984.pdf) formula as in the version of linked papers.
$$_3F_2(1/2,1/2,1/2;1,1;4k^2(1-k^2))=\frac{4K^2(k)}{\pi^2}.\tag{8}$$
Some calculus shows that the combination of both gives:
$$\frac{4K^{2}(k)}{\pi^2}=\frac{1}{\sqrt{1-k^2}}\sum_{n=0}^{\infty}\frac{(2n)!^3}{2^{8n}n!^6}\left(\frac{k^4}{k^2-1} \right)^{n},\tag{9}$$
with $|k|\leq \frac{1}{\sqrt{2}}$. Finally, derivation of $(9)$ and $(8)$ applied to $(6)$ gives:
$$\frac{8K(k)E(k)}{\pi^2}=\frac{2-k^2}{\sqrt{1-k^2}}\sum_{n=0}^{\infty}\frac{(2n+1)(2n)!^3}{2^{8n}n!^6}{\left(\frac{k^4}{
k^2-1}\right)}^n.$$
With $(1)$ and $(9)$ we can now rewrite the $R(k)$ in the question in terms of an $_3F_2$ function and perform integration term by term. The first case in question is $\int_{0}^{k_{1}}R(k)dk$, where $k_{1}=\frac{1}{\sqrt{2}}$ which maybe can be interesting for first instance but the resulting sum involves also Beta functions in the general terms.
Note : Putting $k_{1}=\frac{1}{\sqrt{2}}$ in $(9)$ and $(1)$ we get:
$$\frac{4K^2(k_{1})}{\sqrt{2}\pi^2}=\sum_{n=0}^{\infty}\frac{(-1)^n(2n)!^3}{2^{9n}n!^6}=\frac{\Gamma{(1/4)}^4}{4\sqrt{2}\pi^3},\tag{10}$$
and
$$\frac{8\sqrt{2}E(k_{1})K(k_{1})}{3\pi^2}\\=\sum_{n=0}^{\infty}\frac{(-1)^n(2n)!^3(2n+1)}{2^{9n}n!^6}=\frac{2\sqrt{2}}{3\pi}+\frac{\Gamma(\frac{1}{4})^4}{6\sqrt{2}\pi^3},\tag{11}$$
and we deduce:
$$-\frac{8K^2(k_{1})}{\sqrt{2}\pi^2}+\frac{8\sqrt{2}E(k_{1})K(k_{1})}{\pi^2}\\=\sum_{n=0}^{\infty}\frac{(-1)^n(6n+1)(2n)!^3}{2^{9n}n!^6}=\frac{2\sqrt{2}}{\pi}.\tag{12}$$
Whis is one of famous Ramanujan's like series for $1/\pi$. And playing with some more singular moduli also others can be obtained.
