The completion of a ring with respect to a maximal ideal is always local.
Proof: If $x \in \hat{R}$, then we may write $x = \sum_{i=0}^{\infty} x_i,$ where $x_i \in \mathfrak m^i$. If $x_0 \not\in m,$ then I claim that $x$ is a unit.
Indeed, in this case we may find $y \in R$ such that $x_0 y \equiv 1 \mod m,$
and so $xy = 1 + (x_1y + x_0y - 1) + \sum_{i = 2}^{\infty} x_iy,$ and so it suffices
to show that $\sum_{i=0}^{\infty} x_i$ is a unit under the additional assumption that $x_0 = 1$. But then we can construct an explicit inverse for $x$ using the formula for a geometric series: $x^{-1} = 1 + (x_1 + x_2 + \cdots) + (x_1 + x_2 + \cdots )^2 + \cdots.$
Thus the kernel of the map $x \mapsto x_0 \bmod m$ (i.e. the kernel of the natural projection $\hat{R} \to R/m$) has the property that its complement consists of units, and so it must be the unique maximal ideal of $\hat{R}$,
and so $\hat{R}$ is local. This completes the proof.
In general, if $R$ is a noetherian ring and $\mathfrak m$ a maximal ideal of $R$, then $\hat R$ (the $\mathfrak m$-adic completion of $R$) is a noetherian local ring.
Write $\mathfrak m=(a_1,\dots,a_n)$. Then $\hat R\simeq R[[X_1,\dots,X_n]]/(X_1-a_1,\dots,X_n-a_n)$. A maximal ideal of $R[[X_1,\dots,X_n]]$ has the form $M=\mathfrak n R[[X_1,\dots,X_n]]+(X_1,\dots,X_n)$ with $\mathfrak n\subset R$ a maximal ideal. Since $(X_1-a_1,\dots,X_n-a_n)\subseteq M$ we must have $a_i\in\mathfrak n$ for all $i$, that is, $\mathfrak m=\mathfrak n$. As a consequence we get that $M=\mathfrak m R[[X_1,\dots,X_n]]+(X_1,\dots,X_n)$ and this is the only maximal ideal of $R[[X_1,\dots,X_n]]$ containing $(X_1-a_1,\dots,X_n-a_n)$.
Added in proof. I've found here on page $6$ a more general result: the $I$-adic completion $\hat R$ is quasi-local iff $R/I$ is quasi-local. (Quasi-local means local, but not necessarily noetherian.) In the noetherian case the proof goes exactly as I did before.
Best Answer
Use the fact that completing is the same thing as tensoring with the completed ring, and that the completed ring is flat over the original ring.
(All this is true in view of your hypothesis, of course!)