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 preprint https://arxiv.org/abs/1806.08411 (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 OP's answer, has been proved in https://arxiv.org/abs/1801.02428 several years earlier than in the article of **Tewodros Amdeberhan, Victor Moll, John Lopez Santander, Ken McLaughlin, Christoph Koutschan** https://arxiv.org/abs/2309.02925 . 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.