Suppose $X$ is of finite type (i.e., $\pi _n(X)$ finitely generated for all $n$) and $0$-connected. Then one can use Milnor-type exact sequence to show that
$\pi _*(X_p^{\wedge})\cong \pi _*(X)_p^{\wedge}$
On the other hand, since $X$ is of finite type, so is $\Omega ^{\infty }X$, and according to Bousfield, Kan, {\it Homotopy Limits, Completions, and Localizations} (Splinger Lecture Notes in Mathematics 304}, Part I, Chapter VI example 5.2, we have
$\pi _*((\Omega ^{\infty }X)_p^{\wedge})\cong (\pi _*(\Omega ^{\infty }X))_p^{\wedge}$
(recall that we have finite type hypothesis, so tensoring with $p$-adics amounts to completing at $p$) where $(\Omega ^{\infty }X)_p^{\wedge}$ denotes the Bousfield-Kan $p$-completion of the space $\Omega ^{\infty }X$. It is also known (loc.cit) that under this assumption, Bousfield-Kan completion agrees with QUillen's or Sullivan's profinite $p$-completion.
Combining all these and the isomorphism
$\pi _*(Y)\cong \pi _*(\Omega ^{\infty}Y)\mbox{ for $*\geq 0$)}$ for any spectra $Y$, we see that $(\Omega ^{\infty }X)_p^{\wedge}$ and $\Omega ^{\infty }(X_p^{\wedge})$ are weakly equivalent.
(if you prefer, one can say that the natural map $\Omega ^{\infty }X \rightarrow \Omega ^{\infty }(X_p^{\wedge})$ factors through $(\Omega ^{\infty }X)_p^{\wedge}$ and the resulting map is a weak equivalence.