Let me add one more argument:
For $n \ge 2$:
Suppose the entries in the $n \times n$ matrix $A$ are all distinct. Then there are $(n^2)!$ distinct permutations of $A$.
There are $n!$ row-permutations of $A$ (generated by premultiplication by various permutation matrices), and $n!$ col-permutations of $A$ (generated by post-multiplication by permutation matrices). If we consider all expressions of the form
where $R$ and $C$ each range independently over all $n!$ permutation matrices, we get at most $(n!)^2$ possible results. But for $n > 1$, we have
(n!)^2 &= [ n \cdot (n-1) \cdots 2 \cdot 1 ] [ n \cdot (n-1) \cdots 2 \cdot 1 ] \\
&< [ 2n \cdot (2n-1) \cdots (n+2) \cdot (n+1) ] [ n \cdot (n-1) \cdots 2 \cdot 1 ] \\
&= (2n)! \\
because $2n \le n^2$ for $n \ge 2$, and factorial is an increasing function on the positive integers. So the number of possible results of applying row- and col-permutations to $A$ is smaller than the number of possible permutations of the elements of $A$. Hence there's some permutation of $A$ that does not appear in our list of all $RAC$ matrices.
BTW, just to close this out: for $1 \times 1$ matrices, the answer is "yes, all permutations can in fact be realized by row and column permutations." I suspect you knew that. :)
PS: Following the comment by @Jack M, I want to make clear why it's OK to consider only things of the form $RAC$. Why do we do the column permutations first, and then the rows? (Or vice-versa, if you read things the other way). What if interleaving row and column permutations does something funky? The answer is that if you do a bunch of row ops interleaved with a bunch of column ops, you get the same thing as if you do all the row-ops first, and then all the column ops afterwards (although the row ops have to come in the same order in this rearrangement, and similarly for the column ops). That requires a little proof, but nothing hairy.
What if we do more than one row-permutation, say, two of them? Don't we have to look at $R_1 R_2 A C$ instead? Answer: $R_1 R_2$ will again be a permutation matrix, so we can really consider this as being $(R_1 R_2) A C$, i.e. the matrix I've called $R$ is the product of any number of permutation matrices. And it will always be a matrix with one $1$ in each row and each column. So my counting of possible row-permutations is still valid.