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).
It's certainly not sufficient for the condition to hold for some path; for one, the line integral of any vector field along a constant path is $0$, but not every vector field is conservative.
On the other hand, we have:
Theorem A vector field $\bf F$ (on say, some open set) is conservative iff the line integral of a vector field $\bf F$ over every closed curve in the domain of $\bf F$ is $0$.
The forward implication is a consequence of the F.T.C. for line integrals. We can prove the converse by constructing an explicit potential $f$: If we fix a point $p$ in the domain, for every point $q$ define
$$f(q) := \int_{\gamma} {\bf F} \cdot d{\bf s},$$
where $\gamma$ is any path from $p$ to $q$; this is well-defined (i.e., independent of the path $\gamma$ chosen) because the integral over a closed curve is zero. All that remains to check is that $\nabla f = {\bf F}$ as claimed (see Independepnce of path in a closed curve line integral).
Best Answer
In general, the gradient vector field $\nabla f$ on a regular surface with parametrization $\textbf{x}(u, v)$ and first fundamental form $E=\textbf{x}_u\cdot\textbf{x}_u$, $G=\textbf{x}_v\cdot\textbf{x}_v$ and $F=\textbf{x}_u\cdot\textbf{x}_v=0$ is $$\frac{f_u}{E}\textbf{x}_u+\frac{f_v}{G}\textbf{x}_v.$$ As suggested by the other answer here, we parametrize the sphere with spherical coordinates $\textbf{x}(u, v)=(\sin u\cos v, \sin u\sin v, \cos u)$. Then $E=1$, $G=\sin^2u$ and $F=0$. Expressing the given vector field in terms of $u$ and $v$ and equating it with $\displaystyle f_u\textbf{x}_u+\frac{f_v}{\sin^2u}\textbf{x}_v$, we have $$\begin{cases} f_u&=-\cos u(\sin u+\sin u\sin^2v)\\ f_v&=-\sin^2u\sin v\cos v\end{cases}$$ We then have a desired potential function $\displaystyle f=-\frac{\sin^2u\sin^2v+\sin^2u}{2}$. In terms of $x$ and $y$, $$f=-\frac{x^2+2y^2}{2}.$$.