**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][1].  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 $\operatorname{Sha}(L_i,E)=0$.


  [1]: http://www.warwick.ac.uk/staff/J.E.Cremona//ftp/data/