Let $a_n=\frac{1}{16^n}\binom{2n}n^2$. We have $$ \sum_{n\ge 1} a_n(2H_{2n}-H_n)k^{2n}=-\frac{1}{\pi}K(k)\log(1-k^2). $$ Here $K(x)=\frac{\pi}{2}\sum_{n\ge 0}a_nx^{2n}$ is complete elliptic integral of the first kind. Now devide this identity by $k$ and integrate from $0$ to $1$: \begin{align} \sum_{n\ge 1} \frac{a_n}{2n}(2H_{2n}-H_n)&=-\frac{1}{\pi}\int_0^1K(k)\log(1-k^2)\frac{dk}{k}\\ &=-\frac{1}{2\pi}\int_0^1\frac{\pi}{2}\left(1+\sum_{n\ge 1}a_nx^n\right)\log(1-x)\frac{dx}{x}\\ &=\frac{\pi^2}{24}+\frac14\sum_{n\ge 1}\sum_{m\ge 1}\frac{a_n}{m(n+m)}=\frac{\pi^2}{24}+\frac14\sum_{n\ge 1} \frac{a_n}{n}H_n. \end{align} Thus $$ \sum_{n\ge 1} a_n\frac{4H_{2n}-3H_n}{n}=\frac {\pi^2}{6}.\tag{1} $$ The general series $\, _3F_2(1)$ satisfies $3$-term transformation formula (see Gasper and Rahman, eq. (3.1.3)) \begin{align} \, _3F_2\left({a,b,c\atop d,e};1\right)=\frac{\Gamma (1-a) \Gamma (d) \Gamma (e) \Gamma (c-b)}{\Gamma (c) \Gamma (b-a+1) \Gamma (d-b) \Gamma (e-b)}\, _3F_2\left({b,b-d+1,b-e+1\atop b-c+1,b-a+1};1\right)\\ +\frac{\Gamma (1-a) \Gamma (d) \Gamma (e) \Gamma (b-c)}{\Gamma (b) \Gamma (c-a+1) \Gamma (d-c) \Gamma (e-c)}\, _3F_2\left({c,c-d+1,c-e+1\atop c-a+1,c-b+1};1\right) \end{align} from which one can deduce by setting $e=1$, dividing both sides by $c$ and taking the limit $c\to 0$ $$ \sum_{n\ge 1}\frac{(a)_n(b)_n}{n!(d)_n n}+\psi (1-a)+\psi (b)-\psi (d)-\psi (1)=-\frac{\Gamma (1-a) \Gamma (d)}{b \Gamma (-a+b+1) \Gamma (d-b)}\times\, _3F_2\left({b,b-d+1,b\atop b+1,-a+b+1};1\right),\tag{2} $$ where $\psi$ is digamma function.
Now we apply Newton's method. We have $$ \left\{\frac{d(a)_n}{da}\right\}_{a=1/2}=(1/2)_n\cdot\sum_{k=0}^{n-1} \frac{1}{2k+1}=(1/2)_n(H_{2n}-H_n/2),$$ $$ \left\{\frac{d(a)_n}{da}\right\}_{a=1}=n!\cdot\sum_{k=1}^{n} \frac{1}{k}=n!H_n.$$ First we differentiate (2) wrt to $a$ at $a=b=1/2$, $d=1$ and obtain after simplifications $$ \sum_{n \ge 1}a_n\left(\frac{H_n}{n}+\frac{4H_{2n}-2H_n}{2n+1}\right)=-\frac {\pi^2}{6}+\frac{16 C \log 2}{\pi},\tag{3} $$ where $C$ is catalan's constant.
Similarly by differentiating (2) wrt to $d$ and simplifications we get $$ \sum_{n \ge 1}a_n\left(\frac{2H_{2n}-H_n}{n}+\frac{2H_{n}}{2n+1}\right)=\frac {\pi^2}{2}-\frac{16 C \log 2}{\pi}.\tag{4} $$ Taking the sum of (3) and (4) we get $$ \sum_{n \ge 1}a_n\left(\frac{2H_{2n}}{n}+\frac{4H_{2n}}{2n+1}\right)=\frac {\pi^2}{3}. $$Comparing the last equation and (1) we finally get $$ \sum_{n \ge 1}a_n\left(\frac{2H_{2n}}{n}+\frac{4H_{2n}}{2n+1}\right)=2\cdot \sum_{n\ge 1} a_n\frac{4H_{2n}-3H_n}{n}, $$ which is equivalent to OP's conjecture.
Edit (2019): In the article (see page 20) it was proved that $$ \sum_{n \ge 1}a_n\frac{H_n}{n}=-\frac{5\pi^2}{3}+\frac{64}{\pi}\,\text{Im}\,\text{Li}_3(\tfrac{1+i}{2})+\frac{32}{\pi}C\log 2-2\log^22. $$ Thus equations (1), (3) and (4) allow one to find closed form expressions for the remaining three series $\sum_{n \ge 1}a_n\frac{H_{2n}}{n}$, $\sum_{n \ge 1}a_n\frac{H_n}{2n+1}$, $\sum_{n \ge 1}a_n\frac{H_{2n}}{2n+1}$.
EDIT (2024): The formula $$ \sum_{n=1}^\infty\binom{2n}{n}^2\frac{H_n}{16^n}k^{2n}=K(\sqrt{1-k^2})+\frac{1}{\pi}K(k)\log\frac{k^2}{16(1-k^2)},\tag{1} $$ that was provided in the comment to the question, has been proved in the article several years earlier than in the article of Tewodros Amdeberhan, Victor Moll, John Lopez Santander, Ken McLaughlin, Christoph Koutschan. Moreover, proved in a more general form, and in a cleaner way by finding the generating function of the sequence $\frac{(a)_n(1-a)_n}{(n!)^2}H_n$, where $a$ is a continuous parameter.