Not a complete answer but a bit too long for a comment. Since we consider determinants I assume $R$ is commutative. Let $x \in R$ and consider the matrix
$$X = \left[ \begin{array}{cc} 1 + x & -x \\ x & 1 - x \end{array} \right]$$
which by construction has determinant $1$. By hypothesis, at least one entry of this matrix must be a unit, hence either $x$ is a unit, $1 + x$ is a unit, or $1 - x$ is a unit.
This is very close to one of the characterizations of local rings, that for every $x \in R$ either $x$ or $1 - x$ is a unit. It's not hard to show that this characterization is equivalent to the following: if $\sum x_i = 1$ then at least one of the $x_i$ is a unit. Using this characterization it's not hard to see that the desired property holds for all local rings, since if $X$ is a matrix over a local ring such that $\det X = u$ is a unit then, after multiplying the Laplace expansion of $\det X$ by $u^{-1}$, we see that at least one term in the Laplace expansion must be a unit. So we even conclude that at least $n$ entries of $X$, in distinct rows and columns, must be units.
I'd guess that there ought to be an argument in the other direction showing that any ring with this property is local but I'm not quite familiar enough with local rings to see it.