(I am assuming all entries in the matrix are scalar.)
Lets see if $p(x|y,z) = p(x|z)$. Using the laws of conditional distribution for the normal distribution as mentioned here,
$$p(x|y,z) \sim N\left(0 + [0 \, \, \, \, \, \Sigma_{xz}]\left[ \begin{array}{cc}\Sigma_{yy} & \Sigma_{yz} \\ \Sigma_{yz} & \Sigma_{zz} \end{array} \right]^{-1} \left[ \begin{array}{c} y \\ z \end{array} \right], \Sigma_{xx} - [0 \, \, \, \, \, \Sigma_{xz}]\left[ \begin{array}{cc}\Sigma_{yy} & \Sigma_{yz} \\ \Sigma_{yz} & \Sigma_{zz} \end{array} \right]^{-1} \left[ \begin{array}{c} 0 \\ \Sigma_{xz} \end{array} \right] \right)$$
and
$$p(x|z) = N\left(0 + \Sigma_{xz}\Sigma_{zz}^{-1}(z), \Sigma_{xx} - \Sigma_{xz}\Sigma^{-1}_{zz}\Sigma_{xz} \right) $$
Now
\begin{align*}
[0 \, \, \, \, \, \Sigma_{xz}]\left[ \begin{array}{cc}\Sigma_{yy} & \Sigma_{yz} \\ \Sigma_{yz}^T & \Sigma_{zz} \end{array} \right]^{-1} \left[ \begin{array}{c} y \\ z \end{array} \right] & = [0 \, \, \, \, \, \Sigma_{xz}]\dfrac{1}{\Sigma_{yy}\Sigma_{zz} - \Sigma_{yz}^2}\left[ \begin{array}{cc}\Sigma_{zz} & -\Sigma_{yz} \\ -\Sigma_{yz} & \Sigma_{yy} \end{array} \right] \left[ \begin{array}{c} y \\ z \end{array} \right]\\
& = \dfrac{1}{\Sigma_{yy}\Sigma_{zz} - \Sigma_{yz}^2} \left[\begin{array}{cc}-\Sigma_{xz}\Sigma_{yz} & \Sigma_{xz}\Sigma_{yy}\end{array} \right]\left[ \begin{array}{c} y \\ z \end{array} \right]\\
& \ne \Sigma_{xz}\Sigma_{zz}^{-1}(z).
\end{align*}
(Hopefully, I haven't made any typos.)