With some effort, Mathematica evaluates this as
$$\int_{0}^{1} \int_{0}^{1} \frac{[\ln(1+x^2)-\ln(1+y^2)]^2 }{(x-y)^{2}}dx dy=2 \sqrt{2} \; _4F_3\left(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2};\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2};\tfrac{1}{2}\right)+4 \pi C-C \ln 2-2 i \,\text{Li}_3\left(\tfrac{1}{2}+\tfrac{i}{2}\right)+2 i \,\text{Li}_2\left(1-\tfrac{1+i}{\sqrt{2}}\right) \ln 2-2 i \,\text{Li}_2\left(-\tfrac{1+i}{\sqrt{2}}\right) \ln 2-\left(\tfrac{69}{8}-\tfrac{35 i}{32}\right) \zeta (3)-\tfrac{23 \pi ^3}{192}+\left(\tfrac{7}{2}-\tfrac{7 i}{8}\right) \pi +\tfrac{1}{24} i \ln ^3 2-\tfrac{7}{16} \pi \ln ^2 2-\left(\tfrac{1}{12}+\tfrac{9 i}{32}\right) \pi ^2 \ln 2+\tfrac{1}{2} \pi \ln \left(1+\tfrac{1+i}{\sqrt{2}}\right) \ln 2+\left(\tfrac{7}{8}+\tfrac{11 i}{8}\right) \ln 5-\left(\tfrac{3}{2}+\tfrac{7 i}{4}\right) \arctan\left(\tfrac{1}{2}\right)-\left(\tfrac{11}{2}-\tfrac{7 i}{2}\right) \arctan(1+i)-\tfrac{17}{4} \arctan 2=0.572532$$ with $C$ Catalan's constant and $\text{Li}_n$ the polylog.