Equlibrium for a non-spontaneous reaction
Note: This first part isn't actually that related to the specific scenario you are giving, but I think it's a misconception that you had that led you to ask this question. If you want a direct answer, go to the last paragraph.

Just because it is non-spontaneous, it doesn't mean that no reaction will occur. In fact, the reaction will always occur to some extent, even if only 0.001% of the particles react.

Consider this formula, which I'm sure you have seen before:  $$\Delta G=-RT\ce{ln}K$$
This makes sense because if the reaction is spontaneous ($\Delta G&lt;0$), $K&gt;0$, so the forward reaction, which is spontaneous is favored over the reverse reaction; vice versa.

If we, for example, say $\Delta G = +100$ kJ, at 298 K, $K$ would equal $3.4\cdot10^{-17}$. Thus, the forward reaction pretty much completely dominates the reverse reaction, so we say that the reaction is essentially irreversible. However, the reverse reaction is still happening, to a very small extent. 

If we happened to make $\Delta G = +10$ kJ, $K$ would equal only about $56$, so the reverse reaction becomes visible. The terms spontaneous and non-spontaneous are only generalizations; it doesn't mean that a non-spontaneous reaction doesn't occur at all. There will always be some particles in that substance that will overcome the activation energy and react. 

In terms of your question, since the reaction $\ce{A-&gt;B}$ is non-spontaneous, the reaction $\ce{B-&gt;A}$ is actually spontaneous (think about their equilibrium constants). Thus, the reverse reaction will be favored until the system reaches equilibrium, or when $Q=K$.