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.