This is a corrected version of my original response, incorporating a nice argument by Fedor Petrov.

Hua in his book (cf. review of MR0124306) proved that there are integers $s,K,N>0$ such that every $n>N$ with $n\equiv s\pmod{K}$ is a sum of $s$ $k$-th powers of primes. For any $t>0$ let $M(t)$ denote the set of residues modulo $K$ which can be represented by a sum of $t$ $k$-th powers of primes. Clearly $M(t+1)$ contains $M(t)+p^k$ for any prime $p$, hence $|M(t+1)|> |M(t)|$ unless $M(t+1)$ equals $M(t)+p^k$ for any prime $p$. In this case $M(t)$ is invariant under the shift of $p^k-q^k$ for any two distinct primes $p$ and $q$. The shifts are coprime (e.g. $p^k-q^k$ is coprime with $q$), hence $M(t)=M(t)+1$, and $M(t)$ contains all residues modulo $K$. This argument shows that $|M(K)|=K$, i.e. modulo $K$ every residue class is a sum of $K$ $k$-th powers of primes. If $p$ denotes the largest of the $K^2$ primes used in the latter representation, and $M$ equals $N+Kp^k$, then we have the following. For every $m>M$ there is a sum of $K$ $k$-th powers of primes, denote it by $m'$, such that $m-m'\equiv s\pmod{K}$ and $m-m'>N$. By Hua's theorem, $m-m'$ is a sum of $s$ $k$-th powers of primes, hence in fact every $m>M$ is a sum of $s+K$ $k$-th powers of primes.

To summarize: the statement in Fedor Petrov's original question follows from Hua's theorem.