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, 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, 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 for the definition of $Ш^1_\omega(G,M)$ ). By the answer of Kasper Andersen $Ш^1_\omega(G,M)\ne 0$. 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.)