Understanding efficient classical simulation of quantum computing
Brief (negative) answer

This claim is incorrect as soon as you have three (or more) qubits, as shown by the GHZ paradox, briefly described below, which shows a 3-partite Bell inequality which is beaten by a state stabilized by Pauli operators, which can be prepared and measured by Clifford operations. This paradox is named after Greenberger, Horne and Zeilinger, who found it in 1989 (2007 arXiv reprint here).

The point is that the Gottesman-Knill theorem does not care about locality, but on simplicity. The Gottesmann-Knill theorem (wiki, arXiv) gives a polynomial hidden variable theory for a state constructed with Clifford gates. Nothing prevents this theory to be nonlocal, as the GHZ state shows, but its complexity is bounded : a $n$-qubit state is described by $2n²$ binary hidden variables.

On the other hand Bell inequalities characterize local hidden variable theories, without taking care of their complexity. The violation of a Bell inequality by the GHZ states means that it cannot be described by a model where each player has access to a different local hidden variable.

Details of the GHZ paradox

The paradox can be explained by the following game, with 3 cooperating players and a referee. the referee asks a question, either $X$ or $Y$ to each player, which answers ±1 without communicating with the other players. The referee either asks $X$ to all the players or asks $X$ to one of them and $Y$ to the others. The players win if the product of their answers obeys the equality below corresponding to the question asked:
$$\begin{align}
XXX&amp;=+1\\XYY&amp;=-1\\YXY&amp;=-1 \\YYX&amp;=-1
\end{align}$$
It is easy to see that the above set of equation is not consistent, and tha no local hidden variable theory allows to win the game with a probabiliyt greater than 3/4.

On the other hand, if each of the player has a qubit of a GHZ state $|\mathrm{GHZ}\rangle:=(|000\rangle+|111\rangle)/\sqrt2$, they can win with certainty. Indeed, $|\mathrm{GHZ}\rangle$ is stabilized by the following Pauli operators :
$$\begin{align}
S_0&amp;=+XXX\\S_1&amp;=-XYY\\S_2&amp;=-YXY \\S_3&amp;=-YYX.
\end{align}$$
Therefore, measuring the Pauli operators corresponding to the referee’s question allows the players to always win the game, and therefore violate a Bell inequality with certainty. Furthermore, this state can be prepared from Clifford operations, which invalidates the idea expressed in the paragraph you cite.