Is the blue area greater than the red area?

The four numbered areas are congruent.

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[Added later] The figure below is from a suggested edit by @TomZych, and it shows the congruent parts more clearly. Given all the upvotes to the (probably tongue-in-cheek) comment “This answer also deserves the tick for artistic reasons,” I’m leaving my original “artistic” figure but also adding Tom’s improved version to my answer.

I think sketching the two identical triangles marked with green below makes this rather intuitive. This could also be turned into a formal proof quite easily.

Note that for equal angles $\angle A'OB' = \angle AOB = 90^\circ$, when we subtract a common part $\angle A'OB$ from both sides, we have $\angle AOA' = \angle BOB'$, so the red and cyan triangles are congruent: $\triangle AOA' \cong \triangle BOB'$.

That implies their areas are equal, and when we add a common part $\triangle A'OB$ we get area of the $AOB$ triangle equal to the area of the $A'OB'B$ quadrilateral. Finally, the area of the two squares' common part is constant, independent on the square's rotation angle.