Do all non-conservative vector fields (in 2-space) have corresponding surfaces that are periodic or discontinuous?
No. Non-conservative vector fields can be produced through many other vector potentials. By Helmholtz decomposition, a smooth vector field $F$ can be decomposition into a conservative vector field plus a rotation of some other conservative field:
$$
F = \nabla \phi + \nabla^{\perp} \psi,
$$
where $\nabla^{\perp}$ is like embedding the the 3D curl operator for scalar function in 2D:
$$
\boldsymbol{C}^{1}(\mathbb{R}^2) \hookrightarrow \boldsymbol{C}^{1}(\mathbb{R}^3),
\\
\nabla^{\perp} \psi(x,y) : = \left(\frac{\partial \psi}{\partial y},-\frac{\partial \psi}{\partial x}\right)\mapsto \left(\frac{\partial \psi}{\partial y},-\frac{\partial \psi}{\partial x},0\right) = \nabla\times (0,0,\psi).
$$
Ignoring the conservative part of $F$, we can produce all sorts of non-conservative part of $F$ in $\mathbb{R}^2$ using very "smooth" potential $\psi$, neither periodic nor discontinuous. For example: let $\psi = e^{-x^2-y^2}/2$
$$
F = \nabla^{\perp}\psi = (- y\psi, x\psi).
$$
You can easily check the field you gave is $\nabla^{\perp} xy$, a rotation of the conservative vector field $(x,y)$.
In fact, a $90^{\circ}$ degree rotation of any conservative vector field in $\mathbb{R}^2$ will make it non-conservative.
The surface corresponding to the vector field $F= (y,-x)$ is continuous but periodic, spiraling along the $z$-axis.
As joriki pointed out in the comments, the vector field generated by the spiral you gave is similar to "the gradient field of the polar angle"
$$
F = \nabla \arctan \left(\frac{y}{x}\right) = \left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right). \tag{1}
$$
If the domain contains a curve winding around the origin, then this is not conservative. Otherwise, it is conservative indeed. Like you did there, we let $z$ be parametrized so that we glue different branches of $\arg (x+iy)$ together, the gradient flow is non-conservative. For a more detailed discussion you could refer to my answer here. Roughly the summary is:
$$
\text{zero curl} + \text{simply-connectedness of the domain} \implies \text{conservative}
\\
\text{gradient} + \text{no singularities in the domain} \implies \text{conservative}
$$
Notice "curl zero" means $0$ everywhere, not like (1), if you include $\{0\}$ to make the domain simply-connected, then the curl is zero except this very point.
Lastly, to address your question again, the non-conservative field you found using that potential ("spiral") is actually a special kind among other non-conservative fields. It is a gradient, but it is not conservative (integral around a close path is non-zero).
One can use Green's Theorem if the offending point, say $\vec r_0$, is excluded from the region of integration.
To do that, we deform the boundary contour with a "keyhole" contour that encircles the excluded point $\vec r_0$.
This reduced the problem to evaluating the line integral
$$\lim_{\epsilon \to 0}\int_0^{2\pi}\vec F(\vec r_0+\epsilon \hat r)\cdot \hat \phi \epsilon \, d\phi \tag 1$$
For example, suppose $\vec F(\vec r)=\frac{-\hat xy+\hat yx}{x^2+y^2}$.
NOTE:
The vector $\vec F$ here represents the static magnetic field $\vec H$ (modulo the constant $I/2\pi$) from a line current $I$ aligned along the $z$-axis.
Clearly, for all $(x,y)\ne (0,0)$, the first partial derivatives of $\vec F$ satisfy the relationship
$$\frac{\partial F_y}{\partial x}=\frac{\partial F_x}{\partial y} \tag 2$$
NOTE:
We remark that $(2)$ is the $z$ component of the curl of $\vec F$, which by Maxwell's Equation for static magnetic fields is $\hat z\cdot \nabla \times \vec H=0$ for $\vec r \ne 0$.
However, $\vec F$ is singular at the origin. We can still use $(2)$ in Green's Theorem to evaluate the line integral of $\vec F$ on any contour $C$ that encloses the origin by using the contour deformation given in $(1)$. Proceeding we find with $\vec r_0=0$
$$\begin{align}
\oint_C \vec F(\vec r)\cdot \,d\vec \ell&=\lim_{\epsilon \to 0}\int_0^{2\pi}\vec F(\vec r_0+\epsilon \hat r)\cdot \hat \phi \epsilon \, d\phi\\\\
&=\lim_{\epsilon \to 0}\int_0^{2\pi} \left(\frac{\hat \phi}{\epsilon}\right)\cdot \hat \phi \epsilon \,d\phi\\\\
&=2\pi
\end{align}$$
as expected from Ampere's Law (after multiplying by $I/2\pi$).
Best Answer
The existence of non-conservative vector fields with zero curl is a topological matter, i.e. it only depends on the topology of the space on which the vector fields is defined. This is studied by de Rham cohomology using the language of differential forms.
In your case, the vector field is defined on $R^2$ but, because it is singular at the origin, the domain is actually $R^2$ with the origin removed. Such a space is topologically non-trivial. However, it can be shown that the vector space of non-conservative, irrotational vector fields on punctured $R^2$ is one dimensional. So the vector field you have written is in the appropriate sense unique, where in the appropriate sense means up to addition of any conservative vector field.
Note that the vector field you have written is basically $\mathrm{d}\theta$ expressed in Cartesian coordinates. If you denote by $\theta_i$ the angle $\arctan((y-y_i)/(x-x_i))$ you will see that $\sum_i \mathrm{d}\theta_i$, where the sum is over an arbitrary collection of points $p_i\in R^2$, has zero curl (think of $\mathrm{d}$ as the gradient operator) but is not exact.
In a more general language, irrotational vector field translates to closed differential form, and conservative vector field translates to exact differential form. Vector fields translates to differential 1-forms (to do this properly you need a metric to be defined on your space). Conservative => irrotational translates to exact=> closed. The space of closed 1-forms modulo the exact 1-form is the first de Rham cohomology group.
The first cohomology group is always trivial for a simply-connected space. In the case of $R^2$ with k holes it is $k$-dimensional.