The quantum gates are preserving the purity of the state. Thus, they can be represented as the unitary matrices acting on the state vectors,
\begin{equation}
|\psi_0\rangle \longmapsto |\psi\rangle=U|\psi_0\rangle,
\end{equation}
Or equivalently, on the density matrix,
\begin{equation}
\rho_0=|\psi_0\rangle\langle\psi_0|\longmapsto \rho_{Q}=U\rho_0 U^\dagger=|\psi\rangle\langle\psi|
\end{equation}
For the quantum CNOT gate in $\{|00\rangle,|01\rangle,|10\rangle,|11\rangle\}$ basis,
\begin{equation}
U=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}
\end{equation}
The state after the Hadamard gate is,
\begin{equation}
\psi_0=\frac{1}{\sqrt{2}}|00\rangle+\frac{1}{\sqrt{2}}|10\rangle,
\end{equation}
which corresponds to the density matrix,
\begin{equation}
\rho_0=\begin{pmatrix}\frac{1}{2}&0&\frac{1}{2}&0\\0&0&0&0\\\frac{1}{2}&0&\frac{1}{2}&0\\0&0&0&0\end{pmatrix}
\end{equation}
After the CNOT gate,
\begin{equation}
|\psi\rangle=\frac{1}{\sqrt{2}}|00\rangle+\frac{1}{\sqrt{2}}|11\rangle
\end{equation}
that corresponds to the density matrix,
\begin{equation}
\rho_Q=\begin{pmatrix}\frac{1}{2}&0&0&\frac{1}{2}\\0&0&0&0\\0&0&0&0\\\frac{1}{2}&0&0&\frac{1}{2}\end{pmatrix}
\end{equation}
However when you replace one of the gates with the classical one the final state becomes mixed. I.e. you can no longer describe it using the state vector, but only by means of the density matrix formalism. Then the transformation of the density matrix is a combination of the projection operators (measurements) and the post-projection unitary operations,
\begin{equation}
\rho_0\longmapsto \rho_C=U_{\sigma_1=0}P_{\sigma_1=0}\rho_0 P_{\sigma_1=0}U_{\sigma_1=0}^\dagger+U_{\sigma_1=1}P_{\sigma_1=1}\rho_0 P_{\sigma_1=1}U_{\sigma_1=1}^\dagger
\end{equation}
I.e. we first measure the state, then rotate it depending on the result.
In our case we can represent them as,
\begin{equation}
P_{\sigma_1=0}=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}, P_{\sigma_1=1}=\begin{pmatrix}0&0&0&0\\0&0&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}
\end{equation}
and we may take $U_{\sigma_1=0}=U_{\sigma_1=1}=U$.
Then the resulting density matrix is different from $\rho_Q$,
\begin{equation}
\rho_C=\begin{pmatrix}\frac{1}{2}&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&\frac{1}{2}\end{pmatrix}
\end{equation}
where the lack of the off-diagonal elements signifies the destruction of the superposition, therefore of the entanglement. Basically it may be interpreted as $|00\rangle$ with 1/2 probability and $|11\rangle$ with 1/2 probability but in the classical sense. There is a correlation but not entanglement.
You may distinguish these two states if you measure not just $\sigma_z$ for both qubits, but e.g. $\sigma_z$ for the first one and $\sigma_x$ for the second one. Of course acting on them with extra gates will also produce different results.
