Is a pseudo-Goldstone boson always a pseudoscalar particle? There are several examples of pseudo-Goldstone bosons which are CP-odd particles, such as the pion, as well as many axion-inspired models. 
If we invert the logic,
Are all pseudo-Goldstone boson of CP-odd type? Or, can they be CP-even too? Is there a known example?
 A: 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.
A: The quantum numbers carried by a Goldstone Boson (GB) are determined by the quantum numbers of the spontaneously broken current $J^\mu$ because of the non-vanishing matrix element 
$$
\langle 0|J^\mu(0)|\pi(p)\rangle = i f p^\mu
$$
which defines the decay constant $f$. This equation can be seen as saying that current can be written as 
$$
J^\mu=f\partial^\mu \pi+\ldots
$$ 
So, if $J^\mu$ is a vector current under Parity then GB is a scalar, whereas if $J^\mu$ is an axial current the $\pi$ is a pseudo-scalar. And more generally, if $J^\mu$ carries a non-trivial representation under some internal group (so that it is actually of the type e.g. $J^\mu_a$ with $a$ some index of a given irrep) so it will the GB: the classic example are the pions that form a triplet of the unbroken $SU(2)_V$ symmetry because the axial current of $SU(2)_A$ transforms in the adjoint of $SU(2)_V$ (given that, schematically,  $[Axial,Vector]\sim Axial$ where the indexes are all in the adjoint). They are also pseudo-scalars because the broken current is the axial one.
But, needless to say, there are plenty of examples where the GBs are scalars and not pseudoscalars. One simple example is the following breaking $SU(2)\rightarrow U(1)$ in this explicit weakly coupled model
$$
\mathcal{L}=\frac{1}{2}(\partial\vec\phi)^2-\lambda(\vec\phi^2-v^2)^2+(y\phi\bar\psi\psi+h.c.)
$$
where $\vec\phi$ is a real triplet of $SU(2)$ whereas the fermion $\psi$ is a singlet, and $y
\in \mathbf{R}$. In this model $\vec\phi$ is thus a scalar under parity (this is selected by the coupling to the fermion). The $SU(2)$ current $J^\mu_a\propto \epsilon_{abc}\partial_\mu \phi_b \phi_c$  (where $\phi_a$ is a component of $\vec\phi$) is thus a vector current that is broken spontaneouly by the VEV of $\vec\phi$ that, say, we choose to point along the 3rd direction $\langle\phi_a=\delta_{a3}v\rangle$: 
$$
J^\mu_1\propto \partial_\mu \phi_2 v+\ldots\,, \qquad J^\mu_2\propto -\partial_\mu \phi_1 v+\ldots\,,\qquad J^\mu_3\propto \partial_\mu \phi_1 \phi_2 +\ldots
$$ 
We see that indeed $J_{1,2}^\mu$ starts linearly in the GB's $\pi_1=\phi_2$ and $\pi_2=-\phi_1$ both with decay constant $f=v$ because of the unbroken $U(1)$ symmetry generataed by the unbroken current $J_3^\mu$ (they form a real doublet under $SO(2)$ or a complex scalar $\pi^{\pm}\propto\pi^1\pm i\pi^2$). Clearly, since $\phi_{1,2}$ were scalars (as opposed to the a pseudoscalar) so are the two GB's $\pi^{1,2}$.
A: The pseudo in "pseudoscalar" refers to behaviour under improper Lorentz transformations.
The pseudo in "pseudo Goldstone" just means it behaves similar, but not exactly like a Goldstone boson (it doesn't have mass zero).
The same word is just used in two different contexts, but there is no relation. So true and pseudo Goldstone bosons can be either C or P even or odd.
