Do these matrix rings have non-zero elements that are neither units nor zero divisors?

As you've demonstrated in 1), the question boils down to when $Ax=0$ has a non-trivial solution. It turns out that this is the case if and only if $\det A$ is a zero divisor. I've written this up in a separate post because it's of interest in its own right: necessary and sufficient condition for trivial kernel of a matrix over a commutative ring. It follows that the answer to your question is yes, $A$ is a zero divisor if and only if $\det A$ is a zero divisor, and thus $R^{n\times n}$ inherits from $R$ the property that all non-zero elements are either units or zero divisors.