I do not consider mixed states produced by a partial trace, since I do not think you are intersted in this case.
Consider $N= N_1+N_2$ copies of a given physical system. The first $N_1$ are prepared in the pure state $\psi_1$, the remaining $N_2$ are prepared in the pure stste $\psi_2$. Now mix the systems, e.g. for photons, collect all photons, simultaneously, in a given circuit.
The mixed state representing the ensemble is
$$\rho:= p_1 |\psi_1\rangle \langle \psi_1| + p_2 |\psi_2\rangle \langle \psi_2| \tag{1}$$
where $$p_i:= N_i/N\:,\quad i= 1,2\:.$$
Suppose that $\langle \psi_1|\psi_2\rangle \neq 0$.
Since $\rho$ is selfadjoint, it can be decomposed along the basis of its eigenvectors
$$\rho = \sum_j q_j |\phi_j\rangle \langle \phi_j|\:.\tag{2}$$
This new decomposition cannot coincide with the previous one because here $\langle\phi_i|\phi_j\rangle =0$ if $i\neq j$.
In principle, we can prepare the ensemble by mixing the pure states $\phi_i$ with the weights $q_i$. So also (2) is a possible practical way to produce the final mixed state $\rho$
Once the mixture has been created. All possible information we can experimentally get from the ensemble is of the form $tr(\rho P)$, where $P$ is any test on the system (orthogonal projector).
So, there is no way to decide how the system was prepared: if according to (1) or (2).
In view of this fact it is difficult to argue that mixed states (prepared as above by a direct simultaneous mixing in a given region of space) are epistemic and pure states are ontic.
