I believe that this is not true. For every integer $g$ that is sufficiently positive, I believe that Henry Pinkham has proved that projective cones in $\mathbb{P}^g$ over canonically embedded, genus $g$ curves in $\mathbb{P}^{g-1}$ are "rigid" in the sense that the only (embedded) deformations arise as cones over deformations of the curve in $\mathbb{P}^{g-1}$. (**Ed.** Actually, that might be due to Mumford, but I am having trouble tracking down a reference.) Assuming this, let $\mathcal{I}\subset \mathcal{O}_{\mathbb{P}^g}$ be the ideal sheaf of such a cone.

Denote by $\mathfrak{m}$ the maximal ideal sheaf in $\mathcal{O}_{\mathbb{P}^g}$ of the vertex of the cone. Consider the ideal sheaves $\mathcal{J}$ such that $$\mathfrak{m}\cdot \mathcal{I} \subset \mathcal{J} \subset \mathcal{I}$$ and such that the quotient $\mathcal{I}/\mathcal{J}$ has length $1$, i.e., the quotient equals the skyscraper sheaf of the vertex. By Max Noether's Theorem, the dimension of the locus of such ideals (for fixed ideal $\mathcal{I}$) equals $(g^2-5g+4)/2$. On the other hand, the locus parameterizing ideal sheaves $\mathcal{G}\subset \mathcal{I}$ of colength $1$ whose cokernel has support disjoint from the vertex has dimension $g$. If $g\geq 8$, it appears that the locus of ideal sheaves $\mathcal{J}$ as above forms an irreducible component of the Hilbert scheme (allowing also the cone to vary).

That is bad news. For each such ideal sheaf $\mathcal{I}$ and ideal sheaf $\mathcal{J}$, consider any other ideal sheaf $\mathcal{J}'\subset \mathcal{I}$ of colength $1$ containing $\mathfrak{m}\cdot \mathcal{I}$. Among polynomials $F_d$ of degree $d$ that are contained in $\mathcal{J}\cdot \mathcal{O}(d)$, it is only one more condition to be contained in $\mathcal{J}\cap \mathcal{J}'\cdot \mathcal{O}(d)$. Thus, varying $\mathcal{J}'$ in a $1$-parameter family, this collection of hyperplanes in $H^0(\mathbb{P}^g,\mathcal{J}\cdot \mathcal{O}(d))$ should sweep out the entire vector space. Therefore, it appears to me that the question above has a negative answer.