Do the particle and anti-particle solutions of the Klein-Gordon equation live in different Hilbert spaces? Do the particle and anti-particle solutions of the Klein-Gordon equation live in different Hilbert spaces?   
Our professor said that this is true because the integral of their respective probability density give the values +1 and -1 respectively, but I can't quite understand why this implies that they live in different Hilbert spaces.  
Note: This question is about relativistic quantum mechanics and not quantum field theory.
 A: The question is not well stated as it stands. 
Mathematically speaking it is maybe meaningless or not very interesting: If $\psi \in \cal H$ and $\phi \in \cal H'$ then we can always say that $\psi, \phi \in \cal H \oplus \cal H'$.  
However it is not physically meaningless. Indeed, the correct interpretation of the question is in my opinion whether or not  one-particle states of a charged field belong to unitary equivalent  irreducible representations of the group of physical symmetries of the system: $U(1) \times SO(1,3)^+$.
The answer to OP's question is positive, because the representation of $U(1)$ has two different generators here in accordance with the sign of the charge. 
In both cases (particle/antiparticle) the one-particle Hilbert space $\cal H_\pm$ is isomorphic to $L^2(\mathbb R^3, dp)$, but the generator of $U(1)$ is $Q^{(+)}=qI$ for the particle in $\cal H_+$ and $Q^{(-)}=-qI$ for the antiparticle in $\cal H_-$. 
The two representations are unitarily inequivalent: There is no unitary operator $U : \cal H_+  \to \cal H_-$ such that $UA^{(+)}U^{-1}= A^{(-)}$
where $A^{(\pm)}$ is the generic self-adjoint generator of the relevant representation of  $U(1) \times SO(1,3)^+$. When $A^{(\pm)}=Q^{(\pm)}$, evidently there is no such $U$ with $UQ^{(+)}U^{-1}=Q^{(-)}$.
There is another viewpoint corroborating my positive answer.  (It would be interesting asking OP's professor about the true meaning of his/her remark). One may argue that the overall one-particle space is $\cal H_+ \oplus \cal H_-$. This is not correct, because the superselection rule of the electric charge takes place: no coherent superposition of states in  $\cal H_+$ and $\cal H_-$ are physically permitted. The space of pure states is represented by the unit vectors of  $\cal H_+ \cup \cal H_-$. There are only either particle states (in $\cal H_+$) or antiparticle states  (in $\cal H_-$).
