You have calculated that $\int_\gamma \frac{(x+y)dx-(x-y)dy}{x^2+y^2} = -2a^2\pi.$ (Remember to write the path over which you integrate as the "lower bound" of the integral! Also, the path you've written is not a circle unless $a = 1$. It's worth graphing for your self.) But you have also calculated that the vector field $\mathbf{F} = \left(\frac{x+y}{x^2+y^2},\frac{-(x-y)}{x^2+y^2}\right)$ is irrotational (it's curl is 0). If the vector field were conservative, then we ought to have $\int_\gamma \frac{(x+y)dx-(x-y)dy}{x^2+y^2} = 0$ without even needing to calculate a potential function. This is a basic property of conservative vector fields. So either your direct calculation of the integral is wrong or the field is irrotational but not conservative.
A vector field can be irrotational but not conservative if it is differentiable over a region which is not simply-connected. In the plane, a region is simply-connected if it has no holes. But in fact, the vector field $\mathbf{F}$ is not continuous at the point $(0,0)$ (why?), so it is certainly not differentiable. The path $\gamma$ circles the point $(0,0)$, so any region containing $\gamma$ contains $(0,0)$. We can conclude that $\mathbf{F}$ is not conservative in any region containing $\gamma$.