Is a pseudo-Goldstone boson always a pseudoscalar particle?

I'm not sure why you would think all (pseudo-)Goldstone bosons have to be CP odd. This would be the result if the spontaneously broken symmetry is a chiral symmetry ($SU(2)_A$ for pions, $U(1)_{\text{PQ}}$ for axions), but of course you can spontaneously break other kinds of symmetries too.

For example, consider a complex scalar field with $$\mathcal{L} = |\partial_\mu \phi|^2 + m^2 |\phi|^2 - \lambda |\phi|^4 + \epsilon (\phi^3 + {\phi^*}^3).$$ We can take this complex scalar to be C even and P even. The parameter $\epsilon$ can be taken small technically naturally, because it is an explicit breaking of the $U(1)$ symmetry $\phi \to e^{i \theta} \phi$. Upon spontaneous symmetry breaking, the phase of $\phi$ is a Goldstone boson, which picks up a small mass due to $\epsilon$, and is CP even. This kind of setup is used in Affleck-Dine baryogenesis.