Relaxing the notion of Nash Equilibrium
No, there is not. Consider a game with two players, Ann and Bob. Both choose such vectors with entries $0$ or $1$ of the form $(a_1,a_2,\ldots,a_J)$ or $(b_1,b_2,\ldots,b_J)$, respectively. If $\sum_{i=1}^J a_i+b_i$ is odd, Ann wins and Bob loses. If the number is even, Ann loses and Bob wins. Clearly, one of them loses in every profile of pure strategies, and the loser can win by changing only one entry of their chosen vector. So there is no pure strategy profile at which the revealed preference inequalities hold.