Andreas has pointed out that in any sufficiently saturated nonstandard model $\mathbb{R}^\ast$, the answer is yes.

Meanwhile, let me point out that if one builds $\mathbb{R}^\ast$, as is commonly done, as the ultrapower of $\mathbb{R}$ (including its higher-order structure) by an ultrafilter on $\mathbb{N}$, then the answer is no. Indeed, in such an $\mathbb{R}^\ast$, there is no hyperfinite cover of the standard reals at all. To see this, suppose that $x$ is a nonstandard hyperfinite set of reals. In the ultrapower, $x$ is represented by a function $f$ from $\mathbb{N}$ to the finite sets of reals, so that $f(k)$ is a finite set of reals. By the Los theorem, the reals $r$ with $r^\ast\in x$ must all have $r\in f(k)$ for almost all $k$. But this is a countable set, so $x$ contains at most countably many reals.

Andreas has pointed out that in any sufficiently saturated nonstandard model $\mathbb{R}^\ast$, the answer is yes.

Meanwhile, let me point out that if, as is commonly done, one builds one's hyperreals $\mathbb{R}^\ast$as the ultrapower of $\mathbb{R}$ by an ultrafilter on $\mathbb{N}$, then the answer is no. Indeed, in such a nonstandard $\mathbb{R}^\ast$, there is no hyperfinite cover of the standard reals at all. To see this, suppose that $x$ is a nonstandard hyperfinite set of reals. In the ultrapower, $x$ is represented by a function $f$ from $\mathbb{N}$ to the finite sets of reals, so that $f(k)$ is a finite set of reals. By the Los theorem, the reals $r$ with $r^\ast\in x$ must all have $r\in f(k)$ for almost all $k$. But this is a countable set, so $x$ contains at most countably many reals.