The best you can get if you take the character theory done over $\mathbb C$ and attempt to redo it over an arbitrary field $\Bbbk$ of characteristic $0$ (the usual character theory does not work in positive characteristic because $p$-dimensional representations could end up with the $0$ character), is that the group ring $\Bbbk G$ as a left module over itself is isomorphic to $\bigoplus_V \frac{\dim V}{\dim Hom_\Bbbk(V,V)^G} V$ where $V$ ranges over all irreducible representations and $Hom_\Bbbk(V,V)^G$ is the space of $G$-linear endomorphisms of $V$. This gives you $|G|=\sum_V\frac{(\dim V)^2}{\dim Hom_\Bbbk(V,V)^G}$.
All of the above follows immediately from noticing that the projection formula $\dim V^G=\frac1{|G|}\sum_{g\in G}\chi_V(g)$ still holds (because that formula says nothing more than that $\frac1{|G|}\chi_\Bbbk^*=\sum_{g\in G}g$ is an idempotent with $1$-dimensional image in the ring $\Bbbk G$), that the characters are additive and multiplicative on direct sums and tensor products, and that if $V$ is a representation with character $\chi_V$, then the dual $V^*$ is a representation with character $\chi_{V^*}(g)=\chi_V(g^{-1})$. Then the 'product' $(\alpha,\beta)=\frac1{|G|}\sum_{g\in G}\alpha(g)\beta^{-1}(g)$ is linear in the first variable, so for irreducible representations $V$ and $W$ we get $(\chi_V,\chi_W)=\frac1{|G|}\sum_{g\in G}\chi_V(g)\chi_W(g^{-1})=\frac1{|G|}\sum_{g\in G}\chi_{W^*\otimes V}(g)=\dim(W^*\otimes_{\Bbbk} V)^G=\dim Hom_\Bbbk(V,W)^G$. Since we still have that a $G$-linear map between irreducibles $V$ and $W$ either has trivial kernel or trivial image (as they are subrepresentations), it follows that if $V\neq W$, then $(\chi_V,\chi_W)=0$.
Then linearity in the first variable tells us that for an arbitrary representation $R$ we have a unique decomposition $\chi_R=\sum_V\frac1{\dim Hom_\Bbbk(V,W)^G}(\chi_R,\chi_V)\chi_V$, which implies that characters of irreducible representations are linearly independent. This establishes that the decomposition of a representation into irreducibles corresponds to the represenation of the character as the sum of irreducible characters. Also, because characters are class functions, and the space of class function is generatd by the linearly independent projections of conjugacy classes, we get that there are at most as many irreducible representations as there are conjugacy classes.
Then the usual observation that for the regular representation $R$ of the group ring $\Bbbk G$ as a left module over itself we have $\chi_R(e)=\dim\Bbbk G=|G|$ and $\chi_R(g)=0$ otherwise (look at the traces of the matrices with basisi $G$ for $\Bbbk G$ over $\Bbbk$) implies that $(\chi_R,\chi_V)=\frac1{|G|}|G|\chi_V^*(e)=\dim V$ as usual and hence
$\chi_R=\sum_V\frac{\dim V}{\dim Hom_\Bbbk(V,W)^G}\chi_V$ which gives us the direct sum decomposition of $\Bbbk G$ that I claimed in the beginning.
Note that when $\Bbbk$ is algebraically closed, then surely $\dim_\Bbbk(V,V)^G=1$ for all irreducible representations $V$ since any endomorphism of $V$ must have an eigenvalue and the eigenspace for that eigenvalue for $G$-linear endomorphism must be all of $V$, i.e. the $G$-linear endomorphism must be scalar multiplication, which gives you $\Bbbk G=\bigoplus_V(\frac{\dim V}1)V$ and hence $|G|=\sum_V(\dim V)^2$.
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.
Best Answer
No, characters of a collection of irreducible representations are distinct and linearly independent. This is Corollary 9.22 in Isaacs book "Character Theory of Finite Groups".
If you drop the irreducibility assumption then, in characteristic $p$, the sum of $p$ copies of any representation has the zero character. Brauer characters, which take values in the complex field even when the representation is in finite chracteristic, were introduced (by Brauer) as a means of distinguishing between any pair of non-isomorphic representations.