What about reindexing and induction? The terms $\frac{1}{(2n-1)\cdots(2n-2k-1)}$ have a nice telescopic structure: by the residue theorem
$$ \frac{1}{(2n-1)(2n-3)\cdots(2n-2k-1)}=(-1)^k\sum_{h=0}^{k}\frac{(-1)^h}{(2n-2h-1)}\cdot\frac{1}{2^{k+1}(2h)!!(2k-2h)!!} $$
equals
$$ \frac{(-1)^k}{2^{2k+1}k!} \sum_{h=0}^{k}\frac{(-1)^h}{(2n-2h-1)}\binom{k}{h}.$$
The natural temptation is now to compute
$$ \sum_{n\geq 1}\frac{H_n}{4^{n}}\binom{2n}{n}\frac{1}{2n-2h-1}$$
through $\frac{-\log(1-z)}{1-z}=\sum_{n\geq 1}H_n z^n$ and $\frac{1}{4^n}\binom{2n}{n}=\frac{2}{\pi}\int_{0}^{\pi/2}\left(\cos\theta\right)^{2n}\,d\theta$, multiply both sides by $(-1)^k \binom{k}{h}$, sum over $h=0,1,\ldots,k$ and finish by invoking Fubini's theorem (allowing to switch the integrals with respect to $d\theta$ and $dz$) and the Fourier series $\sum_{m\geq 1}\frac{\cos(m\varphi)}{m}$ and $\sum_{m\geq 1}\frac{\sin(m\varphi)}{m}$.
The only obstruction is that $\frac{1}{2n-2h-1}=\int_{0}^{1}z^n\left[\frac{1}{2z^{h+3/2}}\right]\,dz$ does not hold unconditionally: we would have been happier in having rising Pochhammer symbols rather than falling ones. On the other hand, reindexing fixes this issue. Since $\binom{2n+2}{n+1} = \frac{2(2n+1)}{n+1}\binom{2n}{n}$, the original series can be written as
$$ \sum_{n\geq 1}\frac{H_n \binom{2n}{n}}{4^n(2n-1)} = \sum_{n\geq 0}\frac{2H_{n+1}\binom{2n}{n}}{4^{n+1}(n+1)}=-\frac{1}{\pi}\int_{0}^{1}\sum_{n\geq 0}\int_{0}^{\pi/2}z^n\left(\cos\theta\right)^{2n}\log(1-z)\,d\theta\,dz $$
or
$$ -\frac{1}{\pi}\int_{0}^{1}\int_{0}^{\pi/2}\frac{\log(1-z)}{1-z\cos^2\theta}\,d\theta\,dz =-\frac{1}{2}\int_{0}^{1}\frac{\log(1-z)}{\sqrt{1-z}}\,dz,$$
clearly given by a derivative of the Beta function. This approach works also by replacing $(2n-1)$ with $(2n-1)\cdots(2n-2k-1)$, you just have to be careful in managing the involved constants depending on $k$.
On second thought, I have just applied the binomial transform in disguise.
