No, most force fields refuse to be conservative. The path-independence is a nontrivial constraint in an arbitrarily small region of space, an arbitrary neighborhood.
If the force field is conservative, it must be
$$\nabla\times \vec F = 0$$
because $F = -\nabla\Phi$. It's clear that the curl of $\vec F$ may be nonzero even if you look at a small neighborhood. For example, assume that $\vec F = (0,x,0)$. The curl of this field is $(0,0,1)$. It's just nonzero, even at the very point $(x,y,z)=(0,0,0)$ or any other point, for that matter. If you draw an arbitrarily small area $dS$ in the $xy$-plane, e.g. a small square in the $z=0$ plane, the integral of $\vec F$ over the boundary of the area will be nonzero so, equivalently, the work will be path-dependent. The work will be proportional to $dS$, namely $1\cdot dS$ in my case, but that's the normal scaling.
To make the force field "overlook" that it's not conservative, you have to look at "so small regions" that the derivatives of $\vec F$ are completely invisible. For example, one may be strict and notice that if you only know $\vec F$ at one point only, there is no provable obstruction that would prevent $\vec F$ from being conservative. However, a fairer description of the situation is that if you only know $\vec F$ at one point, you cannot say whether the field is conservative – instead of saying that it is. If the force field were constant, it would be conservative. But to meaningfully say whether the force field is constant, you have to look at nearby different points, too. And as soon as you may measure some derivatives of $\vec F$, the curl of $\vec F$ may be measured as well and it may be nonzero.
You could discuss forces that are "conservative within some error margin". For example, if $\vec F$ were almost constant in your small neighborhood, there would be a sense in which $|\Delta F|\ll |F|$, and as long as you would be satisfied with the approximation $\vec F+\Delta \vec F\approx \vec F$, you could say that the force field is conservative in that region within the same precision. But this whole conclusion only emerges because we threw the baby out with the bath water. Whether a field is conservative may typically be found by looking at small regions of space. A field is "locally" conservative if $\nabla\times \vec F=0$ at the nearby points. One doesn't need to know the behavior of the force field everywhere.