Let me rephrase your question.
Suppose that $\rho$ is a pure state, i.e., it is written as $$\rho =|\psi\rangle \langle \psi|$$ for some unit vector $\psi$.
Your question is the following.
Q1. Is it possible to find a set of vectors $\phi_1, \ldots, \phi_n$ satisfying $$n>1\:,$$ $||\phi_i||=1$, possibly $\langle \phi_i|\phi_j \rangle \neq 0$ for some $i\neq j$, and numbers $q_1, \ldots, q_n$ with $0< q_i <1$ and $\sum_i q_i =1$, such that
$$|\psi\rangle \langle \psi| = \sum_{i=1}^n q_i|\phi_i \rangle \langle \phi_i|$$
and $|\phi_i \rangle \langle \phi_i| \neq |\phi_j\rangle \langle \phi_j|$
for some $i\neq j$?
As far as I understand you already know the following general result.
THEOREM1. Consider an operator $\rho: H \to H$ where $H$ is a complex Hilbert space and $\rho$ is trace class, non-negative and $tr(\rho)=1$.
Under these hypotheses, $\rho$ is a pure state if and only if $tr(\rho) = tr(\rho^2)$.
As a consequence, since the operator $\sum_{i=1}^n q_i|\phi_i \rangle \langle \phi_i|$ is trace class, non-negative with unit trace, Q1 may be restated as follows.
Q2. Is it possible to find a set of vectors $\phi_1, \ldots, \phi_n$ with $$n>1\:,$$ $||\phi_i||=1$, possibly $\langle \phi_i|\phi_j \rangle \neq 0$ for some $i\neq j$, and numbers $q_1, \ldots, q_n$ with $0< q_i <1$ and $\sum_i q_i =1$, such that
$$tr\left[\left(\sum_{i=1}^n q_i|\phi_i \rangle \langle \phi_i|\right)^2\right] =1$$
and $|\phi_i \rangle \langle \phi_i| \neq |\phi_j\rangle \langle \phi_j|$
for some $i\neq j$?
The answer to Q2 is always negative as soon as $n>1$, and thus
it is not necessary to compute the trace of $\left(\sum_{i=1}^n q_i|\phi_i \rangle \langle \phi_i|\right)^2$, just knowing that $n>1$ is enough to decide that the state $\sum_{i=1}^n q_i|\phi_i \rangle \langle \phi_i|$ is not pure unless $|\phi_i \rangle \langle \phi_i|=|\phi_j \rangle \langle \phi_j|$ for all $i,j$.
The proof is the following. First of all let me introduce the Hilbert-Schmidt scalar product between Hilbert Schmidt operators, and thus trace class operators in particular,
$$(\rho|\rho')_{HS} := tr(\rho^*\rho')\:.$$
The associated norm reads
$$||\rho||_{HS}= \sqrt{tr(\rho^*\rho)}\:.$$
Theorem1 can equivalently be restated as follows.
THEOREM2. Consider an operator $\rho: H \to H$ where $H$ is a complex Hilbert space and $\rho$ is trace class, non-negative and $tr(\rho)=1$.
Under these hypotheses, $\rho$ is a pure state if and only if $||\rho||_{HS}=1$.
Now consider an operator $\rho: H \to H$ of the form
$$\rho = \sum_{i=1}^n q_i \rho_i\tag{0}$$
where $\rho_i := |\phi_i \rangle \langle \phi_i|$ with $\phi_i$ and $q_i$ as in Q2. $\rho$ is trace class, non-negative and we want to check if $||\rho||_{HS}=1$ is possible when $n>1$. This condition is equivalent to saying that $\rho$ is pure.
We can always restrict ourselves to deal with a real vector space of trace class operators, since our trace class operators are self-adjoint and the linear combinations we consider are constructed with real (and non-negative) numbers. The scalar product $(\:|\:)_{HS}$ becomes a standard real (symmetric) scalar product in that real subspace.
The crucial observation is that, as it happens in every real vector space equipped with a real scalar product,
$$\left|\left|\sum_{i=1}^n x_i\right|\right| \leq \sum_{i=1}^n ||x_i||\tag{1}$$
and
"$\leq$" is replaced for "$=$" if and only if $x_i = \alpha_i x$ for some fixed $x$ and non negative numbers $\alpha_i$ where $i=1,\ldots,n$.
In other words,
$$\left|\left|\sum_{i=1}^n q_i \rho_i\right|\right|_{HS} \leq \sum_{i=1}^n ||q_i \rho_i||_{HS}\tag{2}$$
and
"$\leq$" is replaced for "$=$" if and only if $q_i \rho_i = \alpha_i T$ for some fixed $T$ and non negative numbers $\alpha_i$ where $i=1,\ldots,n$.
Since we know that
$$\sum_{i=1}^n ||q_i \rho_i||_{HS}= \sum_{i=1}^n q_i ||\rho_i||_{HS} = \sum_{i=1}^n q_i 1 = \sum_{i=1}^n q_i =1$$
we conclude that If $\rho$ in (0) is pure, then the sign "$\leq$" in (2) is replaced by "$=$", so that $q_i\rho_i = \alpha_i T$ for some fixed operator $T$ and reals $\alpha_i$. Taking the trace of both sides $q_i = \alpha_i tr(T)$ where $tr(T) \neq 0$ because $q_i \neq 0$. Re-defining $T \to \rho_0 := \frac{1}{tr T}T$, we have found that there is a positive trace-class operator $\rho_0$ with unit trace such that $\rho_i= \rho_0$ and furthermore $tr \rho_0^2 = tr \rho_i^2 =1$ so that $\rho_0$ is pure and thus it can be written as $\rho_0 := |\phi_0 \rangle \langle \phi_0|$ for some unit vector $\phi_0$.
Summing up, we have obtained that
if $\rho$ in (0) is pure, then
$|\phi_i\rangle \langle \phi_i|= |\phi_0 \rangle \langle \phi_0|$ for all $i=1,\ldots, n$.
Best Answer
The density operator is hermitian. This means that you can find one orthonormal basis $|\phi_i\rangle$ of eigenvectors for it. By definition this means that there are real numbers $p_i$ such that
$$\rho |\phi_i\rangle = p_i |\phi_i\rangle.$$
Now, it is known that if $f(x)$ is one ordinary function, and $A$ is one hermitian operator with orthonormal basis of eigenvectors $|a_i\rangle$ then we define $f(A)$ on this basis to be
$$f(A)|a_i\rangle=f(a_i)|a_i\rangle,$$
which in turn defines $f$ on the whole Hilbert space, since the $|a_i\rangle$ are a basis.
Now this is how $f(\rho)=\rho \ln \rho$ is defined. In the basis of $\rho$ we have
$$f(\rho)|\phi_i\rangle=p_i \ln p_i |\phi_i\rangle.$$
Now remember you can compute the trace in any basis you want. We compute it in this basis, remembering that $f$ has matrix elements:
$$\langle \phi_j |f(\rho)|\phi_i\rangle = p_i \ln p_i \delta_{ij}.$$
Thus the trace is exactly
$$S(\rho)=-\operatorname{Tr}(\rho\ln \rho)=-\sum_{i}p_i \ln p_i.$$
Now, as a remark, in my opinion things can be thought the other way around. This latter expression for entropy was known before QM, with $p_i$ being the probabilities for the microstates. In order to generalize to Quantum Mechanics, remember that $\rho$ represents one ensemble, so that when you have a mixed state, you don't actualy know the actual microstate. On the other hand $\rho$ encodes the probabilities for the microstates as those $p_i$ above.
Thus one could start with the previous knowledge of what $S$ should be in terms of these $p_i$ and arrive at a general expression involving just $\rho$. In simple terms: $S(\rho)$ is defined to yield this result.