The answer is **no**, since first-order deformations can be obstructed, so that they do not give necessarily global embedded deformations of the subscheme. 

For instance, Mumford gives an example of a smooth surface $X$ containing a curve $C$ such that $h^0(N_{C/X}) \neq 0$ (and so $C$ can be deformed in $X$ infinitesimally to the first order) but $C$ cannot be moved globally; in other words, no effective cicle different from $C$ arises in $X$ from the first-order deformations of $C$.

The reduced subscheme of the Hilbert scheme $\textrm{Hilb}_X^{C}$ in a neighborhood of the point $\xi$ corresponding to $C$ consists only of $\xi$, but the tangent space of $\textrm{Hilb}_X^C$ at $\xi$ is $1$-dimensional. This means that the local ring of  $\textrm{Hilb}_X^C$  at $\xi$ contains nilpotents elements, and shows that the appearance of nilpotents is unavoidable also in entirely "classical" questions of algebraic geometry.

For further details, see Shafarevich's book *Classical algebraic geometry 2: schemes and complex manifolds*, page 111 and Mumford's *Lectures on curves on algebraic surfaces*, Lecture 22.