With some effort, Mathematica evaluates this as

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