The obvious problem here is that because $c > 1$, then $c^n \to \infty$ for $n \to \infty$. So we have:
$$\| (c \cdot T)^n x \| \le \underbrace{|c^n|}_{\to \infty}\cdot \underbrace{\|T^nx\|}_{\to 0}$$
So the only possible way for $\left( |c^n| \cdot \| T^n x \| \right) \to 0$ is, if $\|c^n|$ will grow to $\infty$ slower than $\| T^n x\|$ grows to $0$.
Is there any special type of contractions, for which this would be true? What additional assumptions must I make?
Note, that $x$ is any arbitrary element from the Hilbert space $H$.
The necessary and sufficient condition is that the spectral radius $r(T)$ is less than $1.$ Indeed assume $c^nT^nx\to 0$ for any $x.$ By the uniform boundedness principle the norms $c^nT^n$ are uniformly bounded i.e. $c^n\|T^n|\le m$ for some constant $m.$Thus $$r(T)=\lim \|T^n\|^{1/n}\le c^{-1}<1$$ The converse implication does not require advanced theorems. If $r(T)<1$ then $a=[r(T)+1]/2$ satisfies $r(T)<a<1.$ Hence there exists $n_0$ such that $$ \|T^n\|^{1/n}\le a,\quad n\ge n_0$$ We get $a^{-n}\|T^n\|\le 1,$ for $n\ge n_0,$ which implies $b^nT^n\to 0$ for $1<b<a^{-1}.$
Remark The operator $T$ does not need to be a contraction. Its norm can be larger than $1.$ For example let $\{e_n\}_{n=1}^\infty$ be an orthonormal basis and $Te_1=re_2$ and $Te_n=0$ for $n\ge 2.$ Then $T^2=0$ and $\|T\|=r.$