Your two questions (1) and (2) are really the same thing, and (1) just asks for examples where (2) asks if such examples are possible.
I'm not sure if you know this, but I found that the whole thing made a lot more sense for me when put this way: If the field is conservative, then there must be a function $f(x, y)$ such that $\nabla f = \vec F(x, y)$. Using subscripts to denote partials, $\frac{\partial P}{\partial y} = f_{xy}$ and $\frac{\partial Q}{\partial x} = f_{yx}$. And we know thanks to Young's Theorem that if $P_y=Q_x$ then the two functions must have come from a conservative vector field as described above.
That being said, if I happen to be looking at some non-conservative force field $\vec F$, it may happen such that its $P_y$ and $Q_x$ cancel out over some R, as Tunococ mentioned. This would probably happen due to some coincidence or symmetry of the problem - but as soon as R fails to meet whatever vector field's conditions are for $P_y$ and $Q_x$ cancelling out, you'll end up with a nonzero integral.
If you really must have an example, consider $ \vec F(x, y) = x \hat i+x y\hat j $ Then it follows that $P_y = 0$ and $Q_x = y$. I'll pick R to be the rectangular space defined by all $(x, y): a \le x \le b $ and $-c\le y\le c$ for any $c\ge 0, a\epsilon\mathbb{R}, b\epsilon\mathbb{R},$ given $a. You'll notice I have a nice symmetrical rectangle. $ \int\int_R(Q_x-P_y)dA = \int_{x=a}^b\int_{y=-c}^c(y-0)dydx=0 $ But as soon as you loose that symmetric nature, that integral will not be equal to 0.
EDIT: You clarified that you were looking for any two functions $P_y$ and $Q_x$ such that $ \int\int_R(Q_x-P_y)dA = 0 $ over ANY region R (not a specially selected one) for $P_y\ne Q_x$. If you take your focus off of Green's Theorem for a second, we can see why that is not possible. If we think of the magnitude of curl here ($Q_x-P_y$) as simply functions of $x, y$, then you must agree for the conditions you specified there is a nonzero function $D(x, y)$ which satisfies: $ \int\int_R(Q_x-P_y)dA = \int\int_RD(x, y)dA $ Let's assume for simplicity's sake we're dealing with well-behaved functions. The only way that an integral can equal zero is if
1) All values of $D(x, y)$ are zero, which is untrue.
2) Over the entire R our integral somehow manages to cancel out.
However, you're looking for any R.
Let $S$ be the "special region" such that $ \int\int_SD(x, y)dA=0 $ $S$ can either exist or not exist. If it does not, then the function $D(x, y)$ by definition cannot have any region R which makes the integral zero.
If such an $S$ does exist, then we know that within $S$ there exist some regions $S_+,S_-$ and there could exist a region $S_0$ for which the following properties are true: $ \int\int_{S_+}D(x, y)dA>0 $ $ \int\int_{S_-}D(x, y)dA<0 $ $ \int\int_{S_0}D(x, y)dA=0 $ Such that: $ S_++S_-+S_0 = S $ Do you see why?
Now, if you do, then I propose taking a region $A = S_+$. We know for a fact that $ \int\int_AD(x, y)dA \ne 0 $ I made no assumptions about $D(x, y)$, except for being well-behaved, so there will always be a region I can find for a nonzero function $D(x, y)$ such that the integral does not equal 0.