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One can make use of Simplify with Assumptions

I. Compute the sum

s=Sum[HarmonicNumber[n,5]/n^8,{n,1,Infinity}]
(* -(1/63) π^6 Zeta[7]-13/15 π^4 Zeta[9]-55 π^2 Zeta[11]+644 Zeta[13] *)

II. Make a table of Zeta-functions with even arguments

t=Flatten[Table[{ζ[2n]==Zeta[2n]},{n,0,6}]]
(* {ζ[0]==-(1/2),ζ[2]==π^2/6,ζ[4]==π^4/90,ζ[6]==π^6/945,ζ[8]==π^8/9450,ζ[10]==π^10/93555,ζ[12]==(691 π^12)/638512875} *)

III. Simplify with assumptions

Simplify[s/.{π->x,Zeta[n_]->ζ[n]},Assumptions->t/.{π->x}]
(* -15 ζ[6] ζ[7]-78 ζ[4] ζ[9]-330 ζ[2] ζ[11]+644 ζ[13] *)

Comments:

1. A much cleaner solution would be to use the method of Carl Woll to deactivate the Zetafunction. However, it only works if the sum can be computed in the desired terms, i.e., in terms of Zeta[2n+1] and Zeta[2n]. This seems not to be the case.
2. The present method only replaces the occurrences of $\pi^{2n}$ for $n\ge1$. Zeroth-order term is left unchanged.