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][1] to deactivate the `Zeta`function. 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.

  [1]: https://mathematica.stackexchange.com/questions/181160/stop-the-zeta-function-from-evaluating