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.
Best Answer
The smallest counterexample involving irreducible representations of symmetric groups is the $2$-dimensional irreducible module for $\mathbb{C}S_3$. It can be defined over the integers as the submodule $U = \langle e_2-e_1, e_3-e_1\rangle_\mathbb{Z}$ of the natural integral permutation module $\langle e_1, e_2, e_3 \rangle_\mathbb{Z}$. Then $U \otimes_\mathbb{Z} \mathbb{C}$ is irreducible and affords the ordinary character labelled $(2,1)$. The dual $U^\star = \mathrm{Hom}_{\mathbb{Z}}(U,\mathbb{Z})$ is isomorphic to the quotient of $\langle e_1,e_2,e_3 \rangle$ by the trivial submodule $\langle e_1+e_2+e_3\rangle$. The corresponding homomorphisms $\rho, \rho^\star : S_3 \rightarrow \mathrm{GL}_2(\mathbb{Z})$ are such that $\rho(S_3)$ and $\rho^\star(S_3)$ are conjugate in $\mathrm{GL}_2(\mathbb{C})$ but not in $\mathrm{GL}_2(\mathbb{Z})$.
To prove the final claim: if the representations are $\mathbb{Z}$-equivalent then the modules $U \otimes_\mathbb{Z} \mathbb{F}_3$ and $U^\star \otimes_\mathbb{Z} \mathbb{F}_3$ are isomorphic. The first has a trivial submodule spanned by
$$(e_2-e_1) +(e_3-e_1) = e_1+e_2+e_3;$$
the quotient by this submodule is the sign module. The second is its dual, with the factors in the opposite order. Since both are indecomposable, they are not isomorphic.