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). 364
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 $\pi^{1,2}$. 364