Either it's way too hot, or, if we equip L(F,G) with the operator norm (|A|=sup_{|x|=1} |Ax|), F=G=R^2 will be a counterexample.
It cannot be a Hilbert space because A=((1,0),(0,0)) and B=((0,0),(0,1)) don't satisfy the parallelogram identity |A+B|^2+|A-B|^2=2(|A|^2+|B|^2) valid in any Hilbert space.
Of course, in this particular example, you can use the Frechet norm (identifying the 2x2-Matrices with R^4) and get the same topology on L(F,G) from a scalar product...
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