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).
To begin, the reason the gradient of a scalar-valued function is a vector field is:
$$ \nabla \phi = \langle \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y},\frac{\partial \phi}{\partial z} \rangle $$
which is the formula for a vector field. For each point $p$ in some region of $\mathbb{R}^3$ we assign $\nabla \phi (p)$. This makes $\nabla \phi$ a vector field.
Now, if $\vec{F} = \nabla \phi$ then $\nabla \times \vec{F} = \nabla \times \nabla \phi = 0$. However, the converse is not necessarily true. It is possible to have $\nabla \times \vec{F} =0$ throughout some domain $U \subset \mathbb{R}^3$ and yet there does not exist $\phi$ on all of $U$ such that $\nabla \phi = \vec{F}$. If you want the curl vanishing to imply the existence of the potential function $\phi$ then you also need to have a topological condition on $U$. In particular, if $U$ is simply connected then we can apply Stokes' Theorem to arbitrary surfaces in $U$ and for each surface $S$:
$$ \int_{\partial S} \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = 0.$$
Hence integrals of $\vec{F}$ around loops in $U$ vanish hence $\vec{F}$ is path-independent and then we can prove $\displaystyle \phi(\vec{r}) = \int_{p}^{\vec{r}} \vec{F} \cdot d\vec{r}$ is a valid formula for the construction of a potential in $U$.
In any event, if you are currently taking multivariate calculus and you were just introduced to conservative vector fields then relax. In a week or two these things should all gel together. There are a few moving pieces, but, once you see how they all fit it's really pretty. We have the following equivalence:
Suppose $U$ is an open connected subset of $\mathbb{R}^n$ then the following are equivalent
- $\vec{F}$ is conservative; $\vec{F}=\nabla \phi$ on all of $U$
- $\vec{F}$ is path-independent on $U$
- $\oint_C \vec{F} \cdot d\vec{r} =0$ for all closed curves $C$ in $U$
- (add condition $U$ be simply connected) $\nabla \times \vec{F}=0$ on $U$.
Best Answer
Having a potential function and being conservative are equivalent (under some mild assumptions).
Specifically, if a (continuous) vector field is conservative on an open connected region then it has a potential function.
And "Yes" if a vector field fails to be conservative, it cannot have a potential function.
Here are some notes I posted for one of my classes a few years ago... http://mathsci2.appstate.edu/~cookwj/courses/math2130-fall2009/math2130-Line_Int_notes.pdf
A few notes:
1) I didn't list all assumptions everywhere (for example, I wasn't careful to say that I'm assuming things are continuous where needed).
2) In the notes a vector field which possesses a potential function is called a "gradient" vector field.
3) The relevant theorem is on page 5.