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.
To say anything meaningful here, you should start with a connected algebraic group. Finite groups for example are technically affine algebraic groups, but their representations over $\mathbb{C}$ have a literature of their own.
The key reference, following years of partial exploration, involves connected semisimple algebraic groups: see the 1973 paper by Borel and Tits here. (They also provided a short overview of their initial results in English as early as the Tata Institute conference on algebraic geometry held in Bombay in 1968; the proceedings were published in 1969.) The basic answer in this case is that nothing interesting can be done if the group lives in prime characteristic but you want representations over $\mathbb{C}$. Typically, abstract homomorphisms such as representations are gotten by combining algebraic group morphisms with field embeddings.
[On the other hand, passing from representations of semisimple algebraic groups over $\mathbb{C}$ to analogous groups over an algebraically closed field of prime characteristic is quite fruitful. Here Chevalley pioneered a method based on an integral basis of the Lie algebra, which was soon refined by Steinberg and others. For a modern account, see the second edition of J.C. Jantzen's book Representations of Algebraic Groups, Amer. Math. Soc., 2003.]
ADDED: I should also comment on a couple of other issues. (1) While the main emphasis is most often on finite dimensional representations, as in the Borel-Tits work, it's hard to see an interesting example in which a (say connected) affine algebraic group over an algebraically closed field of prime characteristic acts on an infinite dimensional vector space over $\mathbb{C}$. As in the case of Lie groups, most infinite dimensional representations involve some added structure to the vector space. Are there examples to consider? (2) Algebraic groups which aren't semisimple pose other questions in the finite dimensional case. For example, a unipotent group in prime characteristic $p$ only has elements of $p$-power order, unlike unipotent groups in characteristic 0. So one has to consider Jordan-Chevalley decomposition in the algebraic group (as Borel-Tits and earlier work did), and whether this has any interaction with a hypothetical representation over $\mathbb{C}$. Again I'm doubtful that there are interesting examples to study. It would be helpful to provide more motivation for your line of questioning.
Best Answer
You need two conditions for a field to be a splitting field for a specific irreducible representation (in characteristic zero to begin with): It must contain the character values of the representation. For this there is of course a minimal field, the field generated by those values. However, a splitting field must also split a division algebra and for that there is no unique minimal field.
As a specific example we can take the quaternion group of order $8$ over the rational numbers. The component of the group algebra corresponding to the only faithful representation is a quaternion algebra over $\mathbb Q$ ramified at $\infty$ and $2$ and hence is split by any quadratic imaginary field for which $2$ is either ramified or non-split.
To be more concrete about the quaternion representation, the ordinary real quaternion algebra makes sense over the rationals, denoted $\mathbb H_{\mathbb Q}$. It has $\mathbb Q$-basis 1,i,j,k and the usual multiplication table. Then $\pm\{1,i,j,k\}$ is a multiplicative group that is a copy of the quaternion group $Q$ and therefore we have an algebra map $\mathbb Q[Q]\to\mathbb H_{\mathbb Q}$. It is evidently surjective so that $\mathbb H_{\mathbb Q}$ is one of the factors in the group algebra (more precisely $\mathbb Q[Q]=\mathbb Q^4\times\mathbb H_{\mathbb Q}$, where the first four factors correspond to the four one-dimensional representations). Now, for a field $K$ of characteristic zero $\mathbb H_K$ has a two-dimensional irreducible representation exactly when it is split, i.e., when it is isomorphic to the algebra of $2\times2$-matrices. That is thus exactly the condition for a two-dimensional irreducible representation to exist over $K$. Now, it is well-known that the algebra is split precisely when the reduced norm form ($N(\alpha)=\alpha\overline{\alpha}$) restricted to the purely imaginary quaternions has a non-trivial zero. As $N(ai+bj+ck)=a^2+b^2+c^2$ we get the condition that David mentions in the comments.
The situation in positive characteristic is different. As the group algebra can be defined over the prime field and the Brauer group of finite fields is trivial, one only needs for (the mod $p$) character values to be in the field so in that case there is a minimal field.
Addendum: The question of a minimal field for all irreducible representations has essentially the same answer. First the field has to contain all character values, then there are still division algebras to split (in characteristic $0$). The example of the quaternion group still illustrates the problem, all characters are rational-valued and one must still kill the quaternion algebra for which there is no unique minimal field.