The name of the concept you are looking for is the Schur index. The Schur index is $1$ iff the representation can be realized over the field of values. The Schur index divides the degree of the character.
In your case, the the Schur index is either $1$ or $2$. You can use a variety of tests to eliminate $2$, but for instance:
Fein, Burton; Yamada, Toshihiko, The Schur index and the order and exponent of a finite group, J. Algebra 28, 496-498 (1974). ZBL0243.20008.
shows that if the Schur index was $2$, then $4$ divides the exponent of $G$.
In other words, all of your representations are realizable over the field of values.
Isaacs's Character Theory of Finite Groups has most of this in it, and I found the rest of what I needed in Berkovich's Character Theory collections. Let me know if you want more specific textbook references.
Edit: I went ahead and looked up the Isaacs pages, and looks like textbook is enough here: Lemma 10.8 on page 165 handles induced irreducible characters from complemented subgroups, and shows that the Schur index divides the order of the original character. Taking the subgroup to be the rotation subgroup and the original character to be faithful (or whichever one you need for your particular irreducible when $n$ isn't prime), you get that the Schur index divides $1$. The basics of the Schur index are collected in Corollary 10.2 on page 161.
At any rate, Schur indices are nice to know about, and if Isaacs's book doesn't have what you want, then Berkovich (or Huppert) has just a silly number of results helping to calculate it.
Edit: Explicit matrices can be found too. If $n=4k+2$ is not divisible $4$, and $G$ is a dihedral group of order $n$ with presentation $\langle a,b \mid aa=b^n=1, ba=ab^{n-1} \rangle$, then one can use companion polynomials to give an explicit representation (basically creating an induced representation from a complemented subgroup). Send $a$ to $\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}$, also known as multiplication by $x$. Send $b$ to $\begin{pmatrix}0 & -1\\1 & \zeta + \frac{1}{\zeta}\end{pmatrix}$, also known as the companion matrix to the minimum polynomial of $\zeta$ over the field $\mathbb{Q}(\zeta+\frac{1}{\zeta})$, where $\zeta$ is a primitive $(2k+1)$st root of unity.
Compare this to the more direct choice of $a = \begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}$ and $b = \begin{pmatrix} \zeta & 0\\0 & \frac{1}{\zeta}\end{pmatrix}$. If you conjugate this by $\begin{pmatrix}1 & \zeta \\\zeta & 1\end{pmatrix}$ then you get my suggested choice of a representation.
In general, finding pretty, (nearly-)integral representations over a minimal splitting field is hard (and there may not be a unique minimal splitting field), but in some cases you can do it nicely.
Let me know if you continue to find this stuff interesting. I could ramble on quite a bit longer, but I think MO prefers focused answers.
One method for computing branching rules in favorable situations is to use the Littelmann path model---this has a wiki page
http://en.wikipedia.org/wiki/Littelmann_path_model
In this situation (of semisimple $G$ and with $H$ the Levi subgroup of a parabolic) irreducibles essentially never remain irreducible.
Edit in response to the comment below:
Another situation that sometimes occurs (e.g. $SO_n \subseteq SL_n$) is that $H$ is the fixed point set of an involution $\phi$ of $G$. In this case, given any irreducible $G$ module $V$ we can construct another $G$-module $V^\phi$ by twisting by $\phi$. If $V$ and $V^\phi$ are isomorphic as $G$-modules, then we obtain an involution of $V$ as an $H$-module whose $\pm 1$ eigenspaces are $H$-submodules. This allows one to prove that certain $G$-modules are not irreducible as $H$-modules.
Best Answer
Yes, $F$ is a splitting field of $A$. Put $A_E = A\otimes_F E$ for convenience. Let $S'$ be a simple $A$-module. Then $S'\otimes_F E$ has a simple $A_E$-submodule, which by hypothesis is of the form $S\otimes_F E$ for some simple $A$-module $S$. Note that $$0\neq \hom_{A_E}(S\otimes_F E,S'\otimes_F E)\cong \hom_A(S,S'\otimes_F E)$$. But as an $A$-module, $S'\otimes_F E$ is isomorphic to a direct sum of $[E:F]$ copies of $S'$ (this can be infinite if the field extension is infinite, but that's ok). So $$0\neq \hom_A(S,S'\otimes_F E)\leq \hom_A(S,S')^{[E:F]}$$ (this is equality if $[E:F]<\infty$). So by Schur's lemma, $S\cong S'$ and therefore $S'\otimes_F E$ is simple. It now follows that $F$ is a splitting field because $E$ is one.