With respect to ordinal-definability, the answer is no, this is not consistent. The reason is that if there is any non-OD set, then there is such a set of minimal rank. Let $A$ be the set of all non-OD sets of that rank. This is a (paremeter-free) definable set, with no definable member, and even no ordinal-definable member.
Thus, the assertion "every nonempty ordinal-definable set has an ordinal-definable member" is simply provably equivalent to $V=\text{HOD}$. (And I think there is another MO answer to this effect, from a long time ago.)