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.
EDIT: This is a completely new answer.
I will prove that your specific suggestion of defining a Hilbert class field of an elliptic curve $E$ over $K$ does not work. I am referring to your proposal to take the smallest field $L$ such that the corestriction (norm) map $\operatorname{Sha}(L) \to \operatorname{Sha}(K)$ is the zero map. (I have to assume the Birch and Swinnerton-Dyer conjecture (BSD), though, for a few particular elliptic curves over $\mathbf{Q}$.)
Theorem: Assume BSD. There exists a number field $K$ and an elliptic curve $E$ over $K$ such that there is no smallest field extension $L$ of $K$ such that $\operatorname{Cores} \colon \operatorname{Sha}(L,E) \to \operatorname{Sha}(K,E)$ is the zero map.
Proof: We will use BSD data (rank and order of Sha) from Cremona's tables. Let $K=\mathbf{Q}$, and let $E$ be the elliptic curve 571A1, with Weierstrass equation
$$y^2 + y = x^3 - x^2 - 929 x - 10595.$$
Then $\operatorname{rk} E(\mathbf{Q})=0$ and $\#\operatorname{Sha}(\mathbf{Q},E)=4$. Let $L_1 = \mathbf{Q}(\sqrt{-1})$ and $L_2 =\mathbf{Q}(\sqrt{-11})$. It will suffice to show that the Tate-Shafarevich groups $\operatorname{Sha}(L_i,E)$ are trivial.
Let $E_i$ be the $L_i/\mathbf{Q}$-twist of $E$. MAGMA confirms that $E_1$ is curve 9136C1 and $E_2$ is curve 69091A1. According to Cremona's tables, $\operatorname{rk} E_i(\mathbf{Q})=2$ and $\operatorname{Sha}(\mathbf{Q},E_i)=0$, assuming BSD. Thus $\operatorname{rk} E(L_i) = 0+2=2$ and $\operatorname{Sha}(L_i,E)$ is a $2$-group. On the other hand, MAGMA shows that the $2$-Selmer group of $E_{L_i}$ is $(\mathbf{Z}/2\mathbf{Z})^2$. Thus $\operatorname{Sha}(L_i,E)[2]=0$, so $\operatorname{Sha}(L_i,E)=0$.
Best Answer
The genus class field of an extension $K/F$ is defined to be the largest extension $L/K$ with the following properties:
Thus the quick answer to your question is: the Hilbert class field of $K$ is abelian over ${\mathbb Q}$ if and only if the Hilbert class field of $K$ coincides with its genus class field.
The not-so-quick answer would tell you more about the construction of the genus class field. For abelian extensions of the rationals, the construction is easy: everything you'd like to know should be contained in Frölich's book
Basically you will have to look for the largest abelian extension of ${\mathbb Q}$ with the same conductor as $K/{\mathbb Q}$.