The following answer is for finite groups.
In characteristic zero, the group algebra is semisimple, so there are finitely many simple representations. These representations correspond to the blocks in the decomposition as a product of matrix algebras.
Here is a way to determine the number:
Start with the field $K$, adjoin the $g$th roots of unity ($g = |G|$) to get $L$, and consider the Galois group $\Gamma_K=L/K$. This is a subgroup of the multiplicative group of the integers mod $g$. Then $\sigma_t \in \Gamma_K$ corresponding to $t \in (\mathbb{Z}/g\mathbb{Z})^*$ acts on $G$ by raising $x \in G$ to the $t$-th power. The dimension of the space of class functions constant on $\Gamma_K$-orbits is the number of simple $K$-representations.
As for characteristic p, this is modular representation theory, The number of irreducibles is the number of $p$-regular conjugacy classes (where $p$-regular means the period is prime to $p$), when the field contains the $g$th roots of unity for $g$ the order of the group. See, e.g., Serre's Linear Representations of FInite Groups. My guess is that it should be true even without the assumption on the field being sufficiently large.
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
Concerning
A finite dim'l $k$-algebra $A$ is split provided $\operatorname{End}_A(S) = k$ for every simple $A$-module $S$.
[This terminology is consistent with that used in other contexts -- $A$ is split just in case the reductive quotient of the "unit group" $A^\times$ is a split reductive algebraic group over $k$.]
Suppose that $k$ is perfect and that $A_\ell = A \otimes_k \ell$ is a split $\ell$-algebra for a finite, separable extension $\ell \supset k$. Then we have the following:
To see this, I claim first that $\operatorname{End}_A(S) \otimes_k \ell$ is a split semisimple $\ell$-algebra. Indeed, since $k \subset \ell$ is separable, the $A_\ell$-module $S \otimes_k \ell = S_\ell$ is semisimple, say $S_\ell = \bigoplus_i S_i$ as $A_\ell$-module, where $S_i$ is the $T_i$-isotypic component of $S_\ell$ and where $T_i$ are distinct simple $A_\ell$-modules. Since by assumption $\operatorname{End}_{A_\ell}(T_i) = \ell$, we see that $$\operatorname{End}_A(S) \otimes_k \ell \simeq \operatorname{End}_{A_\ell}(S_\ell) = \prod_i \operatorname{End}_{A_\ell}(S_i)$$ is a product of full matrix algebras over $\ell$. Now observe that the center of a split semisimple $\ell$-algebra is a a split commutative etale $\ell$-algbra $\ell \times \cdots \times \ell$. Assertion $(*)$ now follows.
Apply this to $A = kG$ for a finite group $G$. By Torsten's comment (following Emerton's answer) we may suppose $k$ to be perfect. Then $A \otimes_k \ell$ is split for an Abelian extension $\ell$ of $k$ -- a suitable $\ell$ can be obtained by adjoining to $k$ enough roots of unity. It follows that $Z$ is always Galois over $k$ (for $A = kG$).