Giving an "explicit" description of the Hilbert class field of a number field K (or, more generally, all abelian extensions of K) is Hilbert's 12th problem, and has only been solved for Q and for imaginary quadratic fields. The Hilbert class field H of Q(zeta) will only be contained in a cyclotomic field if H = Q(zeta) itself --- since one can explicitly compute that any abelian extension of Q properly containing Q(zeta) is ramified at some place.
This is a great question. Someone will come along with a better answer I'm sure, but here's a bit off the top of my head:
1) The Hilbert class field of a number field $K$ is the maximal everywhere unramified abelian extension of $K$. (Here when we say "$K$" we really mean "$\mathbb{Z}_K$", the ring of integers. That's important in the language of etale maps, because any finite separable field extension is etale.)
In the case of a curve over $\mathbb{C}$, the "problem" is that there are infinitely many unramified abelian extensions. Indeed, Galois group of such is the abelianization of the fundamental group, which is free abelian of rank $2g$ ($g$ = genus of the curve). Let me call this group G.
This implies that the covering space of C corresponding to G has infinite degree, so is a non-algebraic Riemann surface. In fact, I have never really thought about what it looks like. It's fundamental group is the commutator subgroup of the fundamental group of C, which I believe is a free group of infinite rank. I don't think the field of meromorphic functions on this guy is what you want.
2) On the other hand, the Hilbert class group $G$ of $K$ can be viewed as the Picard group of $\mathbb{Z}_K$, which classifies line bundles on $\mathbb{Z}_K$. This generalizes nicely: the Picard group of $C$ is an exension of $\mathbb{Z}$ by a $g$-dimensional complex torus $J(C)$, which has exactly the same abelian fundamental group as $C$ does: indeed their first homology groups are canonically isomorphic. $J(C)$ is called the Jacobian of $C$.
3) It is known that every finite unramified abelian covering of $C$ arises by pulling back an isogeny from $J(C)$.
So there are reasonable claims for calling either $G \cong \mathbb{Z}^{2g}$ and $J(C)$ the Hilbert class group of $C$. These two groups are -- canonically, though I didn't explain why -- Pontrjagin dual to each other, whereas a finite abelian group is (non-canonically) self-Pontrjagin dual. [This suggests I may have done something slightly wrong above.]
As to what the Hilbert class field should be, the analogy doesn't seem so precise. Proceeding most literally you might take the direct limit of the function fields of all of the unramified abelian extensions of $C$, but that doesn't look like such a nice field.
Finally, let me note that things work out much more closely if you replace $\mathbb{C}$ with a finite field $\mathbb{F}_q$. Then the Hilbert class field of the function field of that curve is a finite abelian extension field whose Galois group is isomorphic to $J(C)(\mathbb{F}_q)$, the (finite!) group of $\mathbb{F}_q$-rational points on the Jacobian.
Best Answer
I would first look at the corresponding local question. So $\mathbf{Q}$ gets replaced by $\mathbf{Q}_p$ (where $p$ is a prime), $K$ becomes a finite extension of $\mathbf{Q}_p$, and $H$ is the maximal abelian unramified extension of $K$ (here the adjective "abelian" is superfluous).
Is it possible for $H$ to be galoisian over $\mathbf{Q}_p$ while $K$ itself fails to be so ? If such is the case, the galoisian closure $\tilde K$ of $K$ over $\mathbf{Q}_p$ will be unramified over $K$, so let's look for a $K$ for which $\tilde K$ over $K$ is ramified (of degree $>1$).
To begin somewhere, let's look for $K$ which have degree $p$ over $\mathbf{Q}_p$. If $p=2$, there are no examples because every quadratic extension is galoisian. But if $p\neq2$, by looking at the parametrisation of degree-$p$ extensions of $\mathbf{Q}_p$ as in arXiv:1005.2016, it is easy to see that such $K$ do indeed exist.