Here is a solution using simple tools
We have
$$\sum_{n=1}^\infty x^nH_n=-\frac{\ln(1-x)}{1-x}$$
Replace $x$ with $-x$ then multiply both sides by $-\frac{\ln(1-x)}{x}$ and use the fact that $-\int_0^1 x^{n-1}\ln(1-x)\ dx=\frac{H_n}{n}$
$$\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n}=\int_0^1\frac{\ln(1-x)\ln(1+x)}{x(1+x)}\ dx$$
$$=\underbrace{\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}\ dx}_{-5/8\zeta(3)}-\underbrace{\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ dx}_{\frac{1}{1+x}=y}$$
$$=-\frac58\zeta(3)-\int_{1/2}^1\frac{\ln\left(\frac{y}{2y-1}\right)\ln y}{y}\ dy=-\frac58\zeta(3)-I$$
$$I=\int_{1/2}^1\frac{\ln^2y}{y}\ dy-\int_{1/2}^1\frac{\ln(2y-1)\ln y}{y}\ dy=\frac13\ln^32-\Re\int_{1/2}^1\frac{\ln(1-2y)\ln y}{y}\ dy$$
$$=\frac13\ln^32+\Re\sum_{n=1}^\infty \frac{2^n}{n}\int_{1/2}^1 y^{n-1}\ln y\ dy=\frac13\ln^32+\Re\sum_{n=1}^\infty\frac{2^n}{n}\left(\frac{\ln2}{n2^n}+\frac{1}{n^22^n}-\frac{1}{n^2}\right)$$
$$=\frac13\ln^32+\ln2\zeta(2)+\zeta(3)-\Re\text{Li}_3(2)=\frac18\zeta(3)-\frac12\ln2\zeta(2)+\frac13\ln^32$$
where we used $\Re\text{Li}_3(2)=\frac78\zeta(3)+\frac32\ln2\zeta(2)$
Plug the result of $I$ we get $$\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n}=\frac12\ln2\zeta(2)-\frac34\zeta(3)-\frac13\ln^32$$
A different way to find $\int\frac{\ln(1-x)\ln(1+x)}{1+x} \ dx$
First, add and subtract $\ln2$ and note that $\int\ln\left(\frac{1-x}{2}\right)\ dx=-\text{Li}_2\left(\frac{1+x}{2}\right)$
$$\int\frac{\ln(1-x)\ln(1+x)}{1+x} \ dx=\int\frac{\ln\left(\frac{1-x}{2}\right)\ln(1+x)}{1+x} \ dx+\ln2\int\frac{\ln(1+x)}{1+x}\ dx$$
$$\overset{IBP}{=}-\ln(1+x)\text{Li}_2\left(\frac{1+x}{2}\right)+\int\frac{\text{Li}_2\left(\frac{1+x}{2}\right)}{1+x}\ dx+\frac12\ln2\ln^2(1+x)$$
$$=-\ln(1+x)\text{Li}_2\left(\frac{1+x}{2}\right)+\text{Li}_3\left(\frac{1+x}{2}\right)+\frac12\ln2\ln^2(1+x)$$
Therefore
$$\small{\int_0^a\frac{\ln(1-x)\ln(1+x)}{1+x} \ dx=\text{Li}_3\left(\frac{1+a}{2}\right)-\text{Li}_3\left(\frac{1}{2}\right)-\ln(1+a)\text{Li}_2\left(\frac{1+a}{2}\right)+\frac12\ln2\ln^2(1+a)}$$