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:
$$-\frac{8K^2(k_{1})}{\sqrt{2}\pi^2}+\frac{8\sqrt{2}E(k_{1})K(k_{1})}{\pi^2}\\=\sum_{n=0}^{\infty}\frac{(-1)^n(6n+1)(2n)!^3}{2^{9n}n!^6}=\frac{2\sqrt{2}}{\pi}.\tag{12}$$
Another way to obtain $(12)$ without using the explicit values of $K(\frac{1}{\sqrt{2}})$ and $E(\frac{1}{\sqrt{2}})$ is using Legendre's relation:
$$\frac{\pi}{2}=2E(\frac{1}{\sqrt{2}})K(\frac{1}{\sqrt{2}})-K^2(\frac{1}{\sqrt{2}})\tag{13}$$
Translating to the language of series $(13)$ using $(1)$ and $(9)$ when $k=\frac{1}{\sqrt{2}}$ gives the right hand of $(12)$.
It's curious that Ramanujan didn't list this sum in his famous paper "Modular equations and approximations to $\pi$". The proof of this sum is attributed to Berndt and Baruah using elliptic functions and Jesús Guillera using "creative telescoping".
But we have showed that this sum is the easiest to obtain because Legendre's relation works fine in this case giving a direct link between $K(k)$ and $E(k)$ when $k=\frac{1}{\sqrt{2}}$.
Bauer's series is a little bit complicated but with Landen's transformation (related to the modular equation of degree 2 in the language of elliptic functions) also Legendre's relation is able to link $K(\sqrt{2}-1)$ and $E(\sqrt{2}-1)$.
