Let me write $C_k$ for unordered configurations and $F_k$ for ordered configurations. The natural map $F_k \to C_k$ is a covering map, so the two spaces have the same higher homotopy groups, hence one is aspherical iff the other is. From now on I'll restrict attention to ordered configurations.

The first problem is that $F_1(S) = S$ itself isn't necessarily itself aspherical (e.g. $S$ might be $S^2$). So let's assume that this is the case, and in particular that $S$ is connected. There is a natural fiber sequence

$$S \setminus \{ k \text{ points} \} \to F_{k+1}(S) \to F_k(S)$$

given by forgetting the last point, and inspecting the long exact sequence in homotopy shows that the spaces $F_k(S)$ are all aspherical iff the spaces $S \setminus \{ k \text{ points} \}$ are all aspherical.

If $S$ is closed and orientable of genus $g \ge 1$ then removing a point has the effect of removing the top $2$-cell, so the result is homotopy equivalent to a wedge of $2g$ circles, and from that point on removing a point has the effect of adding a circle to this wedge. If $S$ is nonorientable then we can apply this same argument to its orientable double cover. Either way, it follows that all of the spaces $S \setminus \{ k \text{ points} \}$ are aspherical.

If $S$ is noncompact then I believe it is homotopy equivalent to a wedge of circles and it should again be true that removing a point has the effect of adding a circle to this wedge, but I'm less confident of this.

So I think the answer is all connected surfaces except $S^2$ and $\mathbb{RP}^2$ (since by uniformization, a connected surface is aspherical iff its universal cover isn't $S^2$).