HINT:
Yes, because the Galois group $Gal(\bar K/K)$ acts transitively on the irreducible components of a irreducible $K$ representation $\rho$ .
It is easy to see that if $\rho'$ is a component then $g \cdot \rho'$ is also, with the same multiplicity. Why transitively? Take $\rho_1$,$\ldots$, $\rho_m$ in the same orbit. The character $\rho_1 + \cdots + \rho_m$ takes values in $K$ (being invariant under the Galois group). Let $d = \dim \rho_i$. Then $d\cdot \sum \rho_i$ can be realized over $K$. Indeed, the space of $G$ translates of $\sum\rho_i$ is the $K$-representation with character $d \sum\rho_i$.
Now, if there were several orbits, then for some convenient $N$ ( a common multiple...) the representation $N\cdot \rho$ would have different decomposition into irreducible $K$-representations, not possible.
$\bf{Added}$ In Serre, Linear representation of finite groups, Chap 12, Lemma 12, basically this idea:
If $K\subset L$, $[L\colon K] = n$, $\rho$ representation over $L$ with character $\chi$, then $Res^L_K \rho$ is a $K$ representation with character $Tr^L_K \chi$. Now, if $\chi$ has values in $K$, it follows that $n \cdot \rho$ can be realized over $K$ ( all char are $0$ for simplicity). (this is easier to see than what i wrote above with $\dim \rho_i$ ). Anyways, what we want is : there is some multiple of the representation that is realizable over $K$. )
Don't have a clean reference for the result, although it should be standard. Maybe Serre has it somewhere in the book.
