No, in general the Hasse principle does not hold. I consider the question in the form of GH from MO. I give a counter-example with a non-normal extension $K/\mathbb{Q}$ of degree 6 (the accepted answer of David Speyer gives a counter-example with a normal extension of degree 16). It suffices to take the concrete Galois extension $L/\mathbb{Q}$ of degree 12 with Galois group $G=A_4$ and with cyclic decomposition groups from [the answer of Jeremy Rouse](http://mathoverflow.net/a/242336/4149), and to take $K=L^H$, where $H$ is a subgroup of $G$ of order 2. 

Consider the homomorphism of $\mathbb{Q}$-tori
$$ 
\varphi\colon\  \mathbb{G}_{m,\mathbb{Q}}\times_{\mathbb{Q}}R_{K/\mathbb{Q}}\mathbb{G}_{m,K} \to R_{K/\mathbb{Q}}\mathbb{G}_{m,K}\,,\quad (q,k)\mapsto qk^2.
$$
We wish to show that $Ш^1(\mathbb{Q},\ker \varphi)\ne 0$.
Set $T=R_{K/\mathbb{Q}}\mathbb{G}_{m,K}/\mathbb{G}_{m,\mathbb{Q}}$.
We write $T[2]$ for the subgroup of elements of order dividing 2 in $T$.
From the short exact sequence 
$$ 1\to \mathbb{G}_{m,\mathbb{Q}}\to\ker\varphi\to T[2]\to 1 $$
we obtain a canonical isomorphism 
$$ Ш^1(\mathbb{Q},\ker\varphi)\overset{\sim}{\to} Ш^1(\mathbb{Q}, T[2]).$$
Write $M=T[2]$. It suffices to show that $Ш^1(\mathbb{Q}, M)\ne 0$.

 We have a reduction $Ш^1(\mathbb{Q},M)= Ш^1(L/\mathbb{Q},M)$, 
see [Sansuc's paper](http://www.digizeitschriften.de/dms/img/?PID=GDZPPN002198746), Lemma 1.1(ii).
Since all the decomposition groups of $L/\mathbb{Q}$ are cyclic, we have  $Ш^1(L/\mathbb{Q},M)=Ш^1_\omega(G,M)$ (see [my question](http://mathoverflow.net/q/241876/4149) for the definition of $Ш^1_\omega(G,M)$ ).  By the [answer of Kasper Andersen](http://mathoverflow.net/a/242019/4149) $Ш^1_\omega(G,M)\ne 0$. Thus  $Ш^1(\mathbb{Q},\ker\varphi)\neq 0$, and our extension $K/Q$ is a counter-example to the Hasse principle.

It would be interesting to construct explicitly an element $a\in K$ for which the Hasse principle fails (i.e., $a=qk^2$ locally, but not globally.)