I think what follows gives an "arithmetic" example of what you're looking for: a variety $X$ over a number field $k$ such that $X$ is $\mathbb{Q}$-factorial but $X_L = X \times_k L$ is not, for some finite extension $L/k$. In other words, $X$ is $\mathbb{Q}$-factorial, but not locally $\mathbb{Q}$-factorial in the étale topology.

Take an elliptic curve $E$ over $k$ which has rank zero over $k$, but positive rank over $L$. Embed $E$ as a plane cubic, and let $X$ be the projective cone over $E$. Then the only Cartier divisors on $X$ are multiples of a plane section. Now there is an isomorphism $\mathrm{Cl}_0(X) \to \mathrm{Cl}_0(E) = \mathrm{Pic}_0(E)$ (Hartshorne, II, Ex. 6.3). So the fact that $E(k)$ has rank zero means that $\mathrm{Cl}_0(X)$ is finite, so $\mathrm{Cl}(X)$ has rank 1, so the quotient $\mathrm{Cl}(X)/\mathrm{Pic}(X)$ is finite. Thus $X$ is $\mathbb{Q}$-Cartier. On the other hand, over $L$, $\mathrm{Cl}(X)/\mathrm{Pic}(X)$ has positive rank. Specifically, let $P$ be a point of infinite order on $E(L)$; then the line over $E$ gives a Weil divisor on $X_L$ no multiple of which is Cartier.

I imagine the same argument, starting from a elliptic surface over $\mathbb{P}^1_\mathbb{C}$, could give a "geometric" example.

I think what follows gives an "arithmetic" example of what you're looking for: a variety $X$ over a number field $k$ such that $X$ is $\mathbb{Q}$-factorial but $X_L = X \times_k L$ is not, for some finite extension $L/k$. In other words, $X$ is $\mathbb{Q}$-factorial, but not locally $\mathbb{Q}$-factorial in the étale topology.

Take an elliptic curve $E$ over $k$ which has rank zero over $k$, but positive rank over $L$. Embed $E$ as a plane cubic, and let $X$ be the projective cone over $E$. Then the only Cartier divisor classes on $X$ are multiples of a plane section. Now there is an isomorphism $\mathrm{Cl}_0(X) \to \mathrm{Cl}_0(E) = \mathrm{Pic}_0(E)$ (Hartshorne, II, Ex. 6.3). So the fact that $E(k)$ has rank zero means that $\mathrm{Cl}_0(X)$ is finite, so $\mathrm{Cl}(X)$ has rank 1, so the quotient $\mathrm{Cl}(X)/\mathrm{Pic}(X)$ is finite. Thus $X$ is $\mathbb{Q}$-Cartier. On the other hand, over $L$, $\mathrm{Cl}(X)/\mathrm{Pic}(X)$ has positive rank. Specifically, let $P$ be a point of infinite order on $E(L)$; then the line over $P$ gives a Weil divisor on $X_L$ no multiple of which is Cartier.

I imagine the same argument, starting from a elliptic surface over $\mathbb{P}^1_\mathbb{C}$, could give a "geometric" example.

I think what follows gives an "arithmetic" example of what you're looking for: a variety $X$ over a number field $k$ such that $X$ is $\mathbb{Q}$-factorial but $X_L = X \times_k L$ is not, for some finite extension $L/k$. In other words, $X$ is $\mathbb{Q}$-factorial, but not locally $\mathbb{Q}$-factorial in the étale topology.

Take an elliptic curve $E$ over $k$ which has rank zero over $k$, but positive rank over $L$. Let $X$ be the projective cone over $E$. Then the only Cartier divisors on $X$ are multiples of a plane section. The fact that $E(k)$ has rank zero shows that every divisor on $X$ over $k$ has a multiple which is Cartier. On the other hand, let $P$ be a point of infinite order on $E(L)$; then the line over $E$ gives a Weil divisor on $X_L$ no multiple of which is Cartier.

I imagine the same argument, starting from a elliptic surface over $\mathbb{P}^1_\mathbb{C}$, could give a "geometric" example.

I've got to go and teach now, but will try to fill in the details later.

I think what follows gives an "arithmetic" example of what you're looking for: a variety $X$ over a number field $k$ such that $X$ is $\mathbb{Q}$-factorial but $X_L = X \times_k L$ is not, for some finite extension $L/k$. In other words, $X$ is $\mathbb{Q}$-factorial, but not locally $\mathbb{Q}$-factorial in the étale topology.

Take an elliptic curve $E$ over $k$ which has rank zero over $k$, but positive rank over $L$. Embed $E$ as a plane cubic, and let $X$ be the projective cone over $E$. Then the only Cartier divisors on $X$ are multiples of a plane section. Now there is an isomorphism $\mathrm{Cl}_0(X) \to \mathrm{Cl}_0(E) = \mathrm{Pic}_0(E)$ (Hartshorne, II, Ex. 6.3). So the fact that $E(k)$ has rank zero means that $\mathrm{Cl}_0(X)$ is finite, so $\mathrm{Cl}(X)$ has rank 1, so the quotient $\mathrm{Cl}(X)/\mathrm{Pic}(X)$ is finite. Thus $X$ is $\mathbb{Q}$-Cartier. On the other hand, over $L$, $\mathrm{Cl}(X)/\mathrm{Pic}(X)$ has positive rank. Specifically, let $P$ be a point of infinite order on $E(L)$; then the line over $E$ gives a Weil divisor on $X_L$ no multiple of which is Cartier.

I imagine the same argument, starting from a elliptic surface over $\mathbb{P}^1_\mathbb{C}$, could give a "geometric" example.