The $\sqrt{|G|}$ bound is true in any characteristic. In fact if $R$ is the
Jacobson radical of $k[G]$, then $k[G]/R\cong\prod_iM_{n_i}(k)$ where $i$ runs over
the irreducible representations and $n_i$ is the degree of the corresponding
representation. This gives $\sum_in_i^2=\dim k[G]/R\leq|G|$. In the modular case we
always have that $R\neq0$ so that in particular we have strict inequality. Also for
every irreducible representation in charateristic $p$ there is an irreducible
representation in characteristic $0$ whose degree is $\ge$ than the degree of
the characteristic $p$ representation: Choose a number field $K$ which is a
splitting field for $G$ and let $R$ be its ring of integers and let further $P$
be a maximal ideal of $R$ dividing $p$. We may filter $K[G]$ by a Jordan-Hölder
filtration $W'_i$ so that $W'_i/W' _{i-1}$ is irreducible. Put $W_i:=W'_i\cap R[G]$
so that in particular $W_i/W_{i-1}$ is $R$-torsion free. Hence reducing modulo
$P$ we get a filtration $\overline{W}_i$ of $k[G]$. This filtration can be
refined to a Jordan-Hölder filtration showing that every irreducible
$k[G]$-module which appears in any Jordan-Hölder filtration of $k[G]$ must
appear in any Jordan-Hölder filtration of some
$\overline{W}_i/\overline{W} _{i-1}$ and thus its degree is $\leq
\dim(\overline{W}_i/\overline{W} _{i-1})=\mathrm{rank}(W_i/W _{i-1})=\dim(W'_i/W' _{i-1})$.
Hence the degree of any $k[G]$-representation is $\le$ the degree of some
$K[G]$-representation. It is rare (but does happen) that
$\overline{W}_i/\overline{W} _{i-1}$ is irreducible in the modular case.