$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\KK{\mathbb{K}}\def\LL{\mathbb{L}}\def\Gal{\mathrm{Gal}}\def\FF{\mathbb{F}}$Here are some examples of $\mathbb{K}$ for which this Hasse principle holds, and for which it does not.
Hasse principle $\KK= \QQ(\sqrt{b})$ a quadratic extension of $\QQ$. Let $\LL$ be the normal closure of $\KK(\sqrt{a})$ over $\QQ$. We break into cases according to $\Gal(\LL/\QQ)$.
Case 1 $\Gal(\LL/\QQ)$ is either $\ZZ/4 \ZZ$ or $D_8$ (the dihedral group of order $8$).
In this case, there is a conjugacy class $\sigma$ in $\Gal(\LL/\QQ)$ of order $4$ whose image in $\Gal(\KK/\QQ)$ is the nontrivial class. By Cebatarov, there are infinitely many primes $p$ of $\QQ$ whose Frobenius class is this $\sigma$.
Let $p$ be such a prime, chosen relatively prime to $2$ and $a$. Then $p \mathcal{O}_K$ is prime, with residue field $\mathbb{F}_{p^2}$ and $a \equiv q k^2 \bmod p \mathcal{O}_K$ for some $q \in \mathbb{F}_p$ and some $k \in \mathbb{F}_{p^2}$. But every element of $\mathbb{F}_p$ is square in $\mathbb{F}_{p^2}$, so $a$ is square in $\mathcal{O}/p \mathcal{O}_K$. We deduce that $p$ splits in $K(\sqrt{a})$ and thus in $\LL$, contradicting our choice of Frobenius class.
Case 2 $\Gal(\LL/\QQ)$ is $(\ZZ/2 \ZZ)^2$.
Then $\LL \cong \QQ(\sqrt{b}, \sqrt{c})$ for some $c \in \QQ$, and we have $K(\sqrt{a}) = K(\sqrt{c})$. But then $a=c k^2$ for some $k \in \KK$, as desired. $\square$.
No Hasse principle If we only formulate the Hasse principle as "for all but finitely many primes", then we have a counterexample whenever $\KK/\QQ$ is Galois of odd degree. The reason is simple: For any $a$ whatever in $\KK$, the condition will be satisfied at any unramified prime of odd characteristic.
Proof: Let $\pi$ be such a prime of $\mathcal{O}_{\KK}$, with residue field $\mathbb{F}_{p^f}$. Note that $f$ divides $[\KK:\QQ]$, so it is odd.
Then $\KK_{\pi}^{\times}/\QQ_p^{\times} \cong \FF_{p^f}^{\times}/\FF_p$ (because the extension is unramified). The quotient has order $p^{f-1} + \cdots + p+1$ which is odd, since $f$ is odd. So every element of $\KK_{\pi}$ is of the form $q k^2$ for $q \in \QQ_p^{\times}$ and $k \in \KK_{\pi}$, and it is easy to use an approximation argument to make $q \in \QQ$.
It is easy to choose $a$ to also work at the archimedean places, the even places and the ramified places, and thus get a counterexample to the whole claim.