The question of how to classify all finite-dimensional representations of $C_n$ over an arbitrary field $F$ can be studied using the structure theorem for finitely-generated modules over a principal ideal domain, in this case $F[x]$. The structure theorem asserts that any finitely-generated module is uniquely a finite direct sum of modules of the form $F[x]/p(x)^r$ where $p \in F[x]$ is irreducible and $r$ is a non-negative integer.
If $T$ is an operator acting on $F^k$ for some $n$, then $F^k$ becomes a finitely-generated module over $F[x]$ with $x$ acting by $T$. $T$ gives a representation of the cyclic group $C_n$ if and only if $T^n = 1$, in which case the summands $F[x]/q(x)^r$ in the decomposition of $F^k$ must have the property that $q(x)^r | x^n - 1$.
If $F$ has characteristic $0$ or has characteristic $p$ and $p \nmid n$, then $x^n - 1$ is separable over $F$, hence $r \le 1$ and $F^k$ is a direct sum of irreducible representations, all of which are of the form $F[T]/q(T)$ where $q$ is an irreducible factor of $x^n - 1$ over $F$.
If $F$ has characteristic $p$ and $p | n$, then writing $n = p^s m$ where $p \nmid m$ we have
$$x^n - 1 = (x^m - 1)^{p^s}$$
from which it follows that $r \le p^s$ (but now it is possible to have $r > 1$). If $r > 1$, then the corresponding representation $F[T]/q(T)^r$ is indecomposable and not irreducible, where $q$ is an irreducible factor of $x^m - 1$ over $F$. The irreducible representations occur precisely when $r = 1$. In other words,
The irreducible representations of $C_{p^s m}$, where $p \nmid m$, over a field of characteristic $p$ all factor through the quotient $C_{p^s m} \to C_m$.
One can also see this more directly as follows. If $V$ is an irreducible representation of $C_{p^s m}$ over a field of characteristic $p$ and $T : V \to V$ is the action of a generator, then
$$T^{p^s m} - 1 = (T^m - 1)^{p^s} = 0.$$
Thus $T^m - 1$ is an intertwining operator which is not invertible, so by Schur's lemma it is equal to zero.
Completely reducible is usually called semisimple, which is to say that it can be written as a direct sum of simples. What is a simple module $V$? One that has only the trivial module $\{0\}$ and itself $V$ as submodules. (It has no proper, non-trivial submodules).
This is analogous to how we exclude the integer 1 from the list of prime numbers, so that we don't write decompositions like $30 = 5 \times 3 \times 2 \times 1 \times 1 \times \cdots \times 1$.
Best Answer
I think $20$ is the smallest degree of a faithful representation of the Rubik cube group, certainly in characteristic $0$ or characteristic coprime to the group order, and probably over any field. As Henning Makholm commented, there exist faithful representations of degree $20$, so we just need to show that this is the smallest degree possible.
The Rubik cube group contains a subgroup $H = H_1 \times H_2$, where $H_1$ and $H_2$ have the structures $H_1 = 2^{11}:A_{12}$ and $H_2 =3^7:A_8$.
Now the only nontrivial proper normal subgroups of $H_1$ are its centre $M$ of order $2$, and an elementary abelian group $N$ of order $2^{11}$. In particular, $M$ is its unique minimal normal subgroup, so a minimal degree faithful representation of $H_1$ must be irreducible. Its restriction to $N$ cannot be homogeneous (since $N$ is abelian but not cyclic), and its homogeneous components are permuted by $A_{12}$, so there must be at least $12$ of them.
So the smallest degree of a faithful representation of $H_1$ is $12$ and similarly it is $8$ for $H_2$. By the theory of representations of direct products, the smallest degree of a faithful irreducible representation of $H$ is $12 \times 8 = 96$. Since $H$ has exactly two minimal normal subgroups, the only way we could improve on that is with a representation with two constituents having different minimal normal subgroups in their kernels, and doing that results in a faithful representation of degree (at least) $20$.