That is definitely not true. Let $X$ be an integral curve that has a single ordinary double point $p$ (and no other singular points). Let $\nu:\widetilde{X}\to X$ be the normalization. The fiber of $\nu$ over $p$ consists of two closed points, $p_1$ and $p_2$.

Let $A$ be $\nu_*\mathcal{O}_{\widetilde{X}}$. Let $n$ be $1$. Then $\text{Quot}^1_A$ admits a "fundamental cycle" morphism, $$\text{FC}:\text{Quot}^1_A \to X.$$
But this is not an isomorphism. In fact $\text{Quot}^1_A$ has two irreducible components, one of which is $\widetilde{X}$, and the second of which is the projective line associated to the two-dimensional vector space $\nu_*\mathcal{O}_{\widetilde{X},p}/\mathfrak{m}_p\nu_*\mathcal{O}_{\widetilde{X},p}$, which is canonically isomorphic to $$ (\mathcal{O}_{\widetilde{X},p_1}/\mathfrak{m}_{p_1} \mathcal{O}_{\widetilde{X},p_1}) \oplus (\mathcal{O}_{\widetilde{X},p_2}/\mathfrak{m}_{p_2} \mathcal{O}_{\widetilde{X},p_2}). $$ This $\mathbb{P}^1$ is attached to $\widetilde{X}$ at the two points $p_1$ and $p_2$, with $p_i$ being attached to the point of $\mathbb{P}^1$ determined by the corresponding summand of the two-dimensional vector space. Also $\text{FC}$ is constant on the $\mathbb{P}^1$, mapping it to $p$. All-in-all, $\text{Quot}^1_A$ is quite different from $X$.