Once again, I apologize for my previous comment! The new formulation of the problem is definitely more fun. Here is one solution which only uses smooth, rational curves.

Let $n$ be an integer, $n\geq 4$ (in fact, $n=2$ and $n=3$ make sense if you use the Kontsevich spaces instead of Hilbert schemes).
Start with $\mathbb{P}^{n+1}\times \mathbb{P}^n$ with homogeneous coordinates $([x_0,\dots,x_n,x_{n+1}],[y_0,\dots,y_n])$. Consider the hypersurface $X$ of bidegree $(1,1)$ which is the zero scheme of $x_0y_0+\dots+x_ny_n$. In fact every smooth $(1,1)$-hypersurface in $\mathbb{P}^{n+1}\times \mathbb{P}^n$ is "conjugate" to this one under an automorphism of $\mathbb{P}^{n+1}\times \mathbb{P}^n$. Consider the projection $\pi_2:X\to \mathbb{P}^n$; this is a $\mathbb{P}^n$-bundle $\textbf{Proj}_{\mathbb{P}^n}\text{Sym}^\bullet(\mathcal{E})$ where $\mathcal{E}$ is the locally free, rank $n+1$, $\mathcal{O}_{\mathbb{P}^n}$-module $\mathcal{O}_{\mathbb{P}^n} \oplus Q$ and $Q$ is the cokernel of the tautological morphism $\mathcal{O}_{\mathbb{P}^n}(-1)\to \mathcal{O}_{\mathbb{P}^n}^{\oplus (n+1)}$. From this you can compute that $H^1(X,T_X)$ is zero; it is the same as proving that $H^1(\mathbb{P}^n,\textit{Hom}_{\mathcal{O}_{\mathbb{P}^n}}(\mathcal{E},\mathcal{E}))$ is zero.

Now consider the Hilbert scheme of (flat families of) curves $C$ in $X$ which have constant Hilbert polynomial $1$ with respect to the invertible sheaf $\pi_1^*\mathcal{O}_{\mathbb{P}^{n+1}}(1)$ and which have Hilbert polynomial $t\mapsto nt+1$ with respect to the invertible sheaf $\pi_2^*\mathcal{O}_{\mathbb{P}^n}(1)$. In other words, $C$ is contained in a fiber of $\pi_1$, and under $\pi_2$ the curve $C$ projects isomorphically to a curve of degree $n$ and arithmetic genus $0$. Consider the smooth parameterized curve $V$ which is the image of the following closed immersion.
$$v:\mathbb{P}^1 \to \mathbb{P}^{n+1}\times \mathbb{P}^n, \ \ [t_0,t_1] \mapsto ([0,0,\dots,0,1],[0,t_0^n,t_0^{n-1}t_1,\dots,\widehat{t_0^kt_1^l},\dots,t_0t_1^{n-1},t_1^n])$$
where $(k,l)$ is any pair of integers with $2\leq k,l\leq n-2$ and $k+l=n$. So the image of $V$ in $\mathbb{P}^n$ is a smooth, rational, degree $n$ curve spanning a $\mathbb{P}^{n-1}$ (in particular, it is not a rational normal curve since it is "linearly degenerate"). Just to remark, since the morphism above is $\mathbb{G}_m$-equivariant for the standard action of $\mathbb{G}_m$ on $\mathbb{P}^1$, it suffices to check that $v$ is a closed immersion at $[1,0]$ and $[0,1]$, and this follows since we included the coordinates $t_0^n$, $t_0^{n-1}t_1$, $t_0t_1^{n-1}$ and $t_1^n$.

The deformation theory of smooth integral curves in smooth varieties is described, e.g., in Theorem II.1.7, pp. 95-96, of *Rational Curves on Algebraic Varieties* by János Kollár. In particular, the Hilbert scheme is a local complete intersection at $[C]$ if its (local) dimension equals the "expected dimension", i.e., $\text{deg}_C(c_1(T_X))+(\text{dim}(X)-3)(1-g(C))$. In this case, that works out to be $(n+1)^2-4$. Inside the Hilbert scheme there is an open subscheme $U$ which parameterizes smooth, integral curves; in particular $[V]$ is contained in $U$.

The claim is that $U$ has precisely two irreducible components each having dimension $n^2+2n-3$, and the Hilbert scheme is reduced at the generic point of each irreducible component. Moreover, $[V]$ is contained in the intersection of the two irreducible components. Assuming the claim, it follows that $U$ is a local complete intersection. Since $U$ is a local complete intersection that is generically reduced, $U$ is everywhere reduced. Since $[V]$ is contained in the intersection of the two irreducible components, $[V]$ is a singular point of $U$.

The first irreducible component is the closure ${U}_1$ in $U$ of the open subset $W_1$ parameterizing smooth curves $C$ such that $\pi_1([C])$ is not $[0,\dots,0,1]$. The point is that $\pi_1:X\to \mathbb{P}^{n+1}$ is a $\mathbb{P}^{n-1}$-bundle away from $[0,\dots,0,1]$. Thus $W_1$ is Zariski locally over $\mathbb{P}^{n+1}\setminus \{[0,\dots,0,1]\}$ equal to a product with fiber isomorphic to the Hilbert scheme of smooth, rational, degree $n$ curves in $\mathbb{P}^{n-1}$. That fiber space is well-described in many places: it is a smooth, rational variety of dimension $n^2+n-4$. Hence $W_1$ is a smooth, rational variety of dimension $(n^2+n-4)+(n+1)=n^2+2n-3$.

The second irreducible component $U_2$ is the Hilbert scheme of smooth, rational, degree $n$ curves in the fiber $D:=\pi_1^{-1}([0,\dots,0,1])\cap X$. Note that $\pi_2:D \to \mathbb{P}^n$ is an isomorphism. Again the Hilbert scheme of smooth, rational, degree $n$ curves in $\mathbb{P}^n$ is a smooth, rational variety of dimension $n^2+2n-3$. The scheme $U$ equals the union of its closed subsets $U_1\cup U_2$, and even $W_1\cup U_2$, from which it follows that $\text{dim}(U)$ equals $\text{dim}(U_1)=\text{dim}(U_2)=n^2+2n-3$. Thus $U$ is indeed a local complete intersection. Moreover, $U_1$ contains the dense open subscheme $W_1$, which is smooth, hence reduced. Thus $U$ is reduced at the generic point of $U_1$.

The last issue is whether or not $U$ is reduced at the generic point of $U_2$. This is the same as the question of whether or not $H^1(C,N_{D/X})$ is zero for a rational normal curve (of degree $n$) in $D \cong \mathbb{P}^n$. In fact $N_{D/X}\cong \pi_2^*Q^\vee$, with $Q$ as above. By an explicit computation, for a rational normal curve $f:\mathbb{P}^1\to \mathbb{P}^n$, the pullback $f^*Q$ is isomorphic to $\mathcal{O}_{\mathbb{P}^1}(1)^{\oplus n}$, and hence $f^*N_{D/X}$ is isomorphic to $\mathcal{O}_{\mathbb{P}^1}(-1)^{\oplus n}$. In particular, $H^1(\mathbb{P}^1,f^*N_{D/X})$ is zero. From this it follows that $U$ is smooth at the generic point of $U_2$.

Finally, why is $[V]$ in the intersection $U_1\cap U_2$? Clearly, $[V]$ is contained in $U_2$. To see that $[V]$ is contained in $U_1=\overline{W}_1$, consider the family, parameterized by the coordinate $s$, of image curves $V_s$ of the following closed immersions.$$v_s:\mathbb{P}^1 \to \mathbb{P}^{n+1}\times \mathbb{P}^n, \ \ [t_0,t_1] \mapsto ([s,0,\dots,0,1],[0,t_0^n,t_0^{n-1}t_1,\dots,\widehat{t_0^kt_1^l},\dots,t_0t_1^{n-1},t_1^n]).$$ For $s\neq 0$, the images are curves in $W_1$. For $s=0$, the image equals the curve $V$. Thus $[V]$ is the image of $0$ under the corresponding morphism $$\tilde{v}:\mathbb{A}^1\to U, \ s\mapsto [V_s],$$ and thus is in the closure of $\tilde{v}(\mathbb{G}_m) \subset W_1$.