Since you are using Mathematica, you definitely want to take a look at the extremely useful package HolonomicFunctions by Christoph Koutschan.
In your particular example,
Annihilator[f1[d], {S[d]}]
shows that this function satisfies the recurrence \begin{equation} (4+4d)f_1(d+4)+(4-7d)f_1(d+2)-(6-3d)f_1(d)=0. \end{equation}
Once known, Mathematica itself can check symbolically that both of your functions satisfy this recurrence:
(4+4d)f1[d+4] + (4-7d)f1[d+2] - (6-3d)f1[d] // FullSimplify
(4+4d)f2[d+4] + (4-7d)f2[d+2] - (6-3d)f2[d] // FullSimplify
After checking initial conditions (which Mathematica can do) it follows that $f_1(d)=f_2(d)$ for all even integers $d$
But as I'm typing I see that the OP just figured all of this out by himself... ;) So let me just mention that one strategy now could be to look at $f_1-f_2$, show that it satisfies the necessary exponential growth conditions (should be alright after combining the poles; apart from these each function seems to be good by itself), and apply Carlson's Theorem. I hope that helps...
