We have
$$ \int_{0}^{1}\log^2(1-x) x^{n-1}\,dx =\frac{d^2}{da^2}\left.\int_{0}^{1}(1-x)^a x^{n-1}\,dx\right|_{a=0}=\frac{H_n^2-H_n^{(2)}}{n}$$
by differentiating Euler's Beta function. Since $\text{Li}_2(-x)=\sum_{n\geq 1}\frac{(-1)^n x^n}{n^2}$, the evaluation of $\mathcal{J}$ boils down to the evaluation of two alternating Euler sums with odd weight:
$$\mathcal{J}=\sum_{n\geq 1}(-1)^n\frac{H_n^2-H_n^{(2)}}{n^3}. \tag{A}$$
Now $\sum_{n\geq 1}\frac{H_n^2-H_n^{(2)}}{n^3}=S_{11,3}-S_{2,3} =8\,\zeta(5)-4\,\zeta(2)\,\zeta(3)$ follows from standard results on cubic Euler sums with weight $5$ (Flajolet and Salvy), and by summation by parts the evaluation of $\sum_{n\geq 1}\frac{H_{2n}}{n^3}$ boils down to the evaluation of $\sum_{n\geq 1}\frac{H_n^{(3)}}{n^2} = \frac{11}{2}\,\zeta(5)-2\,\zeta(2)\,\zeta(3)$ and
$$ \mathcal{K}_1 = \sum_{n\geq 1}\frac{H_n^{(3)}}{(2n-1)^2}=-\int_{0}^{1}\frac{\text{Li}_3(x^2)\log(x)}{x^2(1-x^2)}\,dx.\tag{B}$$
If we manage to compute $\mathcal{K}_1$ and
$$ \mathcal{K}_2 = \sum_{n\geq 1}\frac{H_{2n}^2}{n^3},\tag{C} $$
or just $\int_{0}^{1}\frac{\log^2(1-x)\text{Li}_2(x^{\color{red}{2}})}{x}\,dx$, we are done.
This kind of Euler sums have been studied by Xu, Yang and Zhang. Page $13$ of this recent article proves that $\mathcal{J}$ is a polynomial in $\zeta(2),\zeta(3),\zeta(5),\log(2),\text{Li}_4\left(\frac{1}{2}\right)$ and $\text{Li}_5\left(\frac{1}{2}\right)$.
