Consider the vector field $$\vec{F} = \frac{x}{r}\hat{x} + \frac{y}{r}\hat{y} + \frac{z}{r}\hat{z}$$
with $r = \sqrt{x^2 + y^2 + z^2}$. If I can find a function $V$ such that $\vec{F} = \nabla V$, then the unit radial vector field, by definition, will be called conservative. I find that $V = \sqrt{x^2 + y^2 + z^2}$ works and therefore, $\vec{F}$ is conservative.
Now let's attack the problem using the curl. Can we use the curl to determine if $\vec{F}$ is conservative? Yes, however in addition to the curl being zero, we have to add in the necessary condition that the domain of $\vec{F}$ be simply connected. I find that $\nabla \times \vec{F} = \vec{0}$. Therefore I conclude on any simply connected domain not containing the origin, there is some potential $V$ out there. The largest simply connected domain that I could have could be, for instance, all of $\mathbb{R}^3$ except half the plane $x = 0$ (the full plane $x = 0$ is too much but if we remove just half of it, the domain is still simply connected). Question 1: In my first paragraph, I found a potential function. However, my textbook lacks details on where I can call $\vec{F}$ conservative. It basically says, if you can find a $V$, then $\vec{F} = \nabla V$ is called conservative on it's domain. It's domain is all 3-tuples of real numbers except the origin. So should I call $\vec{F}$ conservative over all of $\mathbb{R}^3$ except the origin? But my second paragraph (this one) says that I should call $\vec{F}$ conservative over any simply connected domains? Which one do I go with?
Question 2: My 2nd issue comes when I look at a counterclockwise vector field divided by $r^2$
$$\vec{G} = \frac{-y}{r^2} \hat{x} + \frac{x}{r^2}\hat{y}$$
I find that $\nabla \times \vec{G} = \vec{0}$. This is good. So over any simply connected domain not containing the origin, I can call $\vec{G}$ conservative and find some potential function for it. Now let's consider for fun
$$ \oint_{\text{unit circle}} \vec{G} \cdot d\vec{s} $$
For 1 revolution, this integral is $2\pi$. For $n$ revolutions, this integral is $2\pi n$. For conservative vector fields, any circulation should always give $0$. This shows us (at least somewhat) why $\vec{G}$ can't be called conservative on domains that contain the origin. But interestingly
$$\oint_{\text{unit circle}} \vec{F} \cdot d\vec{s} = 0 $$
Does this mean, even though the theory says that we call $\vec{F}$ conservative over simply connected domains, the unit radial field is an exception? We can call the unit radial field conservative over all of $\mathbb{R}^3$, including the origin?
Started as a comment, I believe it is strong enough as a statement to turn into a full answer to the question.
The condition states that if the field is irrotational (that is, null curl over its domain), then it is conservative over simply connected domains that are subsets of its domain. I'm omitting all the details, since they are not as important as the point of this answer.
The unit radial field is no exception to the rule, in fact you can check for yourself that it is conservative over simply connected domains. However, the rule says nothing about what the field does over other domains.
Specifically, it does not say that the field must be non conservative over domains that contain the origin.