The 1-categorical version of motivic spaces is locally cartesian closed but not a quasitopos. The idea is that there exists an $\mathbb A^1$-contractible scheme that is covered by $\mathbb A^1$-rigid schemes. This leads to an equivalence relation that is maximally non-effective, and it just happens to be a strong equivalence relation. In particular this argument gives a very simple proof that motivic spaces over an arbitrary nonempty base scheme do not form an ∞-topos.

Let me give a somewhat more general recipe. Suppose $C$ is a small site and $W$ is a collection of morphisms of $C$ that is stable under base change. Let $P_W(C)\subset P(C)$ be the category of presheaves that invert $W$, call them $W$-invariant, and let
$$ Shv_W(C) = Shv(C) \cap P_W(C) \subset P(C) $$
be the category of $W$-invariant sheaves on $C$. Note first that $Shv_W(C)$ is presentable and locally cartesian closed. In fact, the reflector $L: P(C) \to Shv_W(C)$ is semi-left exact (a.k.a. locally cartesian), since it is a transfinite composition of sheafification and $W$-invariantification, and the latter is semi-left exact because $W$ is stable under base change.

Now, assume that there exists a sheaf $X\in Shv(C)$ with $LX\simeq *$ and an effective epimorphism $U\to X$ in $Shv(C)$ with $U$ and $R=U\times_XU$ both $W$-invariant. Assume also that $R\hookrightarrow U\times U$ is not an isomorphism. Then $Shv_W(C)$ is not a topos. Indeed, the equivalence relation $R\rightrightarrows U$ in $Shv_W(C)$ has quotient $LX\simeq *$, so it is not effective.

So if we can find an example where $R\hookrightarrow U\times U$ is a *strong* monomorphism in $Shv_W(C)$, we are done. I don't see a general way to achieve this, but here is an example.

**Example.**

$S$ is a nonempty reduced scheme, $C$ is the category of smooth $S$-schemes with a subcanonical topology finer than Zariski, and $W=\{X\times\mathbb A^1\to X\}$. Take $U\to X$ to be the covering map $$\mathbb A^1\setminus\{0\} \sqcup \mathbb A^1\setminus\{1\}\to \mathbb A^1.$$ Then $R=\mathbb A^1\setminus\{0\} \sqcup\mathbb A^1\setminus\{0,1\} \sqcup\mathbb A^1\setminus\{0,1\} \sqcup \mathbb A^1\setminus\{1\}$. Since every scheme in $C$ is reduced and $\mathcal O^\times$ is $\mathbb A^1$-invariant on reduced schemes, both $U$ and $R$ are $\mathbb A^1$-invariant (representable) sheaves.

I claim that $R\hookrightarrow U\times U$ is a strong monomorphism in $Shv_W(C)$. It suffices to show that the open immersion $\mathbb G_m\setminus\{1\} \hookrightarrow \mathbb G_m$ is a strong monomorphism. It is the kernel of the obvious pair $\mathbb G_m\rightrightarrows D$ where $D$ is $\mathbb G_m$ with doubled $1$ (a smooth non-separated $S$-scheme), so it suffices to show that $D$ is $\mathbb A^1$-invariant. An $S$-morphism $X\to D$ consists of an invertible function $f\in\mathcal O(X)^\times$ and a coproduct decomposition $f^{-1}(1)=U\sqcup V$. So an $S$-morphism $X\times\mathbb A^1\to D$ is an invertible function $f\in\mathcal O(X)^\times$ and a coproduct decomposition of $f^{-1}(1)\times\mathbb A^1$. But any such coproduct decomposition is the pullback of a coproduct decomposition of $f^{-1}(1)$, since $\mathbb A^1$ over a field is connected. QED