First of all you have to distinguish the classical uncertainty form the quantum one. The density matrix can be written as $$\rho=\sum w_i|\alpha_i\rangle\langle\alpha_i|$$ where the $w_i$ can be the classical probabilities, if you can't say exactly where the state is in the Hilbert space, or the quantum ones, if you don't want (or can't) write the state as a definite vector of the Hilbert space.
Classical
As an example take a two level system and consider the case in which you classically can't say in which one of the two states the system is. In that case your density matrix will be $$\rho=\frac{1}{2}|0\rangle\langle0|+\frac{1}{2}|1\rangle\langle1|$$ and your entropy will be $S=-Tr(\rho\log\rho)=\log2$.
This entropy is a measure of your classic uncertainty about the state and has nothing to do with quantum uncertainty.
Quantum
Now consider a system composed as the two two level systems of above each one with his Hilber space $\mathcal{H}_1$ and $\mathcal{H}_2$. If you want to write the total Hilbert space you have to do the tensor product and you will have some vectors that can't be write in the separate Hilbert spaces. A famous example is one of the Bell's states
$$|\psi\rangle=\frac{1}{\sqrt{2}}\left(|0\rangle\otimes|1\rangle+|1\rangle\otimes|0\rangle\right)~.$$
While a total state can be always written in base where his state is one of the base itself and so its density matrix will be like $$\rho=\begin{pmatrix}1&...&0\\.&...&.\\0&...&0\end{pmatrix}$$ and its entropy $\rho=-Tr(\rho\log\rho)=0$, this is not true for a single subsystem of the Bell's state for which the density matrix will be obtained with the trace operator on the Hilber sbace of the other subsystem acting on the total density matrix:
$$\rho_{A}=Tr_B(\rho)=\frac{1}{2}\begin{pmatrix}1&0\\0&1\end{pmatrix}~.$$
In this case the Von Neumann entropy will be $S=-Tr(\rho\log\rho)=\log2$.
This entropy measures the quantum correlations between the two subsystems and are always different from zero if the state you are considering is somehow interacting with something else (another subsystem or an environment).
From the classical and the maximally entangled scenario you can see that there are not differences between the two density matrices and so, given ad arbitary density matrix you do not have a criterion to distinguish the two cases and the Von Neumann entropy will result the same. It can be show that distinguish an entangled or non-entangled density matrix is a NP-hard problem known as Quantum Separability Problem.
