I'd like to expand a bit on Francesco Polizzi's excellent answer, with a couple of examples. He's right of course that deformations may be obstructed, but there's another issue which prevents infinitesimal deformations from being realized as families of rationally equivalent cycles.
As Francesco observes, the existence of obstructed deformations of a subscheme $Z\subset X$ is equivalent to the non-smoothness of the Hilbert $\operatorname{Hilb}_X^{p(Z)}$ at the point corresponding to $Z$. Now suppose one has some unobstructed deformation of $Z$, say a flat family $\mathcal{Z}\to k[t]/t^n$ with closed fiber $Z$. If $X$ is projective and $[Z]\in\operatorname{Hilb}_X^{p(Z)}$ is a smooth point, then by the projectivity of $\operatorname{Hilb}^{p(Z)}_X$ we can slice $\operatorname{Hilb}^{p(Z)}_X$ by hypersurfaces tangent to the family $\mathcal{Z}$ and thus realize $\mathcal{Z}$ inside a family over some curve.
This will give a family of subschemes algebraically equivalent to $Z$--but unless the curve is rational, they may not be rationally equivalent to $Z$. As a simple example, consider deforming points in an elliptic curve. Such deformations are always unobstructed, since elliptic curves are smooth by definition. But no two distinct points in an elliptic curve are rationally equivalent (because e.g. any map from a rational curve to an elliptic curve is constant, or because distinct points on an elliptic curve give distinct line bundles).
The question you are asking is thus not a local question about the Hilbert scheme, but rather one about its global geometry---namely, is there a rational curve in the Hilbert scheme realizing some given germ of a deformation. This is very hard in general, even for zero cycles on surfaces--see e.g. Bloch's conjecture, and the related beautiful paper by Mumford herehere.
