We will assume that $\widehat M$ is a free $\widehat R$-module of rank $n$ with generators $\widehat m_1, \dots, \widehat m_n.$ By this post, there exist generators $m_1, \dots, m_n$ of $M$ such that $m_i + IM$ is the image of $\widehat m_i$ under the canonical surjection $\rho : \widehat M \to M / IM.$ Consequently, there exists a surjective $R$-linear map $\pi : R^n \to M$ that sends $(r_1, \dots, r_n) \mapsto r_1 m_1 + \cdots + r_n m_n.$ We have a short exact sequence $$0 \to \ker \pi \to R^n \to M \to 0$$ of $R$-modules. By hypothesis that $R$ is Zariski, we have that $\widehat R$ is flat as an $R$-module, hence $$0 \to \widehat R \otimes_R \ker \pi \to \widehat R \otimes_R R^n \to \widehat R \otimes_R M \to 0$$ is a short exact sequence of $R$-modules. Of course, we have that $\widehat R \otimes_R R^n \cong \widehat R^n$ and $\widehat R \otimes_R M \cong \widehat M,$ hence the map $\widehat R \otimes_R R^n \to \widehat R \otimes_R M$ can be identified with the map $(\widehat r_1, \dots, \widehat r_n) \mapsto \widehat r_1 \widehat m_1 + \cdots + \widehat r_n \widehat m_n.$ Considering that $\widehat M$ is a free $\widehat R$-module, this map is injective, hence $\widehat R \otimes_R \ker \pi = 0.$ But as $\widehat R$ is faithfully flat over $R,$ we have that $\ker \pi = 0.$
If $M$ is projective, then there is module $K$ s.t. $M \oplus K \cong R^{(\Lambda)}$. For any multiplicative set $S \subset R$, one gets $S^{-1}M \oplus S^{-1} K \cong (S^{-1}R)^{(\Lambda)}$. Hence, any localization of a projective module is projective.
If $(R,m)$ is local and $M$ is finitely generated projective then $M$ is free. Here is a proof. The short exact sequence
$$
0 \to K \to R^n \to M \to 0,
$$
splits, hence $K$ is finitely generated and sequence
$$
0 \to K \otimes R/m \to R^n \otimes R/m \to M \otimes R/m \to 0,
$$
also splits. By Nakayama's lemma we can choose generators of $M$ in such way that the map $R^n \otimes R/m \to M \otimes R/m$ is an isomorphism, then $K \otimes R/m \cong K/mK=0$. By Nakayma's lemma $K=0$, since it is finitely generated.
Fact: if $M$ is finitely presented, then $S^{-1}Hom_R(M,N) \cong Hom_{S^{-1}R} (S^{-1}M, S^{-1}N)$.
To check that $M$ is projective is enough to check that for any surjection $N \to N'$ map $Hom(M,N) \to Hom(M, N')$ is surjective, but this can be checked stalkwise if $M$ is finitely presented (by the fact).
There are plenty of other proofs, but often they involve flat modules, in this proof we only use two ways to characterize projective modules and the fact. Notice that this proof doesn't use that $R$ is Noetherian.
Best Answer
$\newcommand\ideal{\mathfrak}$The completion of a finitely generated module $M$ over a Noetherian ring $A$ can be obtained by extension of scalars: $\hat M\cong\hat A\otimes_AM$. By associativity of tensor product: \begin{align} \hat{\ideal a}\otimes_{\hat A}\hat M &\cong(\hat A\otimes_A\ideal a)\otimes_{\hat A}(\hat A\otimes_AM)\\ &\cong((\hat A\otimes_A\ideal a)\otimes_{\hat A}\hat A)\otimes_AM\\ &\cong(\hat A\otimes_A\ideal a)\otimes_AM\\ &\cong\hat A\otimes_A(\ideal a\otimes_AM) \end{align} From the canonical epimorphism $\ideal a\otimes_AM\twoheadrightarrow\ideal aM$, we get the commutative diagram below, where the top row is surjective:$\require{AMScd}$ \begin{CD} \hat A\otimes_A(\ideal a\otimes_AM)@>>>\hat A\otimes_A(\ideal aM)\\ @V\sim VV@VV\sim V\\ \hat{\ideal a}\otimes_{\hat A}\hat M@>>>\widehat{\ideal aM} \end{CD} Consequently, the bottom row is surjective as well, but since its image is $\hat{\ideal a}\hat M$, this concludes the proof.