In Problem 25.3 this is repeated asking the reader for a proof. I wonder though if this is really true. I know this can be proven for Lie algebras of compact Lie groups (or to be precise, every representation is equivalent to a hermitian one), but I would not know how to show it for semisimple Lie algebras.

page 230 of the book by fuchs and schweigert may be relevant. it shows how to construct a hermitian form on a highest weight module, which in the semi simple case is non degenerate. but i am a novice here.

thank you for your answer. Unfortunately this argument is a bit too abstract for me. However, if this means that the statement is indeed true I will try to find a more elementary proof. If I fail I will ask for help in the Homework & Courseworks forum section.

As you've already mentioned you know the proof for compact Lie groups, i.e. Weyl's unitarian trick, this result is a simple application of it. I'll start with a lightning review of Weyl's trick. To a compact Lie group G, associate a biinvariant measure dg. Now form an inner product on the complex vector space V, being used for the representation. Take this measure and average it over the group.Since V was taken to be complex, this averaged measure must be Hermitian.
Now we consider a complex semi-simple Lie algebra g. There exists a compact form, y, the Lie algebra of the real compact Lie group Y whose complexifiied Lie algebra is g.As a representation of y and g will agree on a complex vector space, we can apply the lifting property of simply connected groups to conclude that a representation of g must also be unitary. Have a nice day.

thank you very much. So for every complex semi-simple Lie algebra there is a real compact Lie group whose complexification is the semi-simple Lie algebra. This is probably a basic fact, but as a Physicist I am struggling to find the relevant knowledge of this subject (I especially wonder if someone having only the information of the book in the OP is able to come up with such a proof...).
A nice day to you too