Disproportionation of Au(II)
I think this could be a counter answer. In contrast, I found a literature evidence that disproportionation of $\ce{Ag(II)}$ to $\ce{Ag(I)}$ and $\ce{Ag(III)}$ has had happens Ref.1). The abstract states that:


  Interaction of $\ce{Ag+}$ salts in anhydrous liquid hydrogen fluoride, $\ce{aHF}$, with $\ce{AgF4-}$ salts gives amorphous red-brown diamagnetic $\ce{Ag^IAg^{III}F4}$, which transforms exothermally to brown, paramagnetic, microcrystalline $\ce{Ag^{II}F2}$ below $\pu{0 ^\circ C}$. $\ce{Ag^IAu^{III}F4}$ prepared from $\ce{Ag+}$ and $\ce{AuF4-}$ in $\ce{aHF}$ has a tetragonal unit cell and a $\ce{KBrF4}$ type lattice, with $a = \pu{5.788(1) Å}$, $c = \pu{10.806(2) Å}$, and $Z = 4$. Blue-green $\ce{Ag^{II}FAsF6}$ disproportionates in $\ce{aHF}$ (in the absence of $\ce{F-}$ acceptors) to colorless $\ce{Ag^IAsF6}$ and a black pseudotrifluoride, $\ce{(Ag^{II}F+)2Ag^{III}F4^-AsF6-}$. The latter and other $\ce{(AgF)2AgF4MF6}$ salts are also generated by oxidation of $\ce{AgF2}$ or $\ce{AgF+}$ salts in $\ce{aHF}$ with $\ce{F2}$ or in solutions of $\ce{O2+MF6-}$ salts ($\ce{M = As, Sb, Pt, Au, Ru}$). Single crystals of $\ce{(AgF)2AgF4AsF6}$ were grown from an $\ce{AgFAsF6/AsF5}$ solution in $\ce{aHF}$ standing over $\ce{AgF2}$ or $\ce{AgFBF4}$, with $\ce{F2}$ as the oxidant. They are monoclinic, $P2/c$, at $\pu{20 ^\circ C}$, with $a = \pu{5.6045(6) Å}$, $b = \pu{5.2567(6) Å}$, $c = \pu{7.8061(8) Å}$, $β = 96.594(9)°$, and $Z = 1$. The structure consists of $\ce{(AgF)_n^{n+}}$ chains ($\ce{F−Ag−F} = 180^\circ $, $\ce{Ag−F−Ag} = 153.9(11)^\circ $, $\ce{Ag−F} = \pu{2.003(4) Å}$), parallel to $c$, that enclose stacks of alternating $\ce{AgF4-}$ and $\ce{AsF6-}$, each anion making bridging contact with four $\ce{Ag(II)}$ cations of the four surrounding chains “caging” them. There is no registry between the ordered array in one “cage” and that in any neighboring “cage”. The $\ce{F-}$ ligand anion bridges between the anions and, with the $\ce{Ag(II)}$ of the chains, generates a trifluoride-like structure. $\ce{(AgF)_2AgF4AsF6}$ [like other $\ce{(AgF)_n^{n+}}$ salts] is a temperature-independent paramagnet except for a Curie “tail” below $\pu{50 K}$.


However, I cannot find a counter example for $\ce{Cu(II)}$ complexes. Yet, $\ce{Cu(I)}$ is known to disproportionate to $\ce{Cu(II)}$ and $\ce{Cu(0)}$, which has limited $\ce{Cu(I)}$ chemistry. For example, if you react $\ce{Cu(I)}$ oxide ($\ce{Cu2O}$) with dilute sulfuric acid (hot), you'd get a brown precipitate of $\ce{Cu(0)}$ and a blue solution of $\ce{CuSO4}$ (a $\ce{Cu(II)}$ salt), because the disproportionation reaction dominates. You would not get the expected copper(I) sulfate solution from this reaction.

$$\ce{Cu2O + H2SO4 -&gt; Cu + CuSO4 + H2O}$$

However, if you use certain ligands to form $\ce{Cu(I)}$ complexes, they would stabilizes the $\ce{Cu(I)}$ oxidation state. For example, both $\ce{[Cu(NH3)2]+}$ and $\ce{[CuCl2]-}$ are $\ce{Cu(I)}$ complexes, which don't disproportionate.

References:


Ciping Shen, Boris Žemva, George M. Lucier, Oliver Graudejus, John A. Allman, Neil Bartlett, "Disproportionation of$\ce{Ag(II)}$ to $\ce{Ag(I)}$ and $\ce{Ag(III)}$ in Fluoride Systems and Syntheses and Structures of $\ce{(AgF+)_2AgF4^-MF6-}$ Salts ($\ce{M = As, Sb, Pt, Au, Ru}$)," Inorg. Chem. 1999, 38(20), 4570-4577 (https://doi.org/10.1021/ic9905603).
Also read for the Disproportionation of$\ce{Au(II)}$: Khaldoon A. Barakat, Thomas R. Cundari, Hassan Rabaâ, Mohammad A. Omary, "Disproportionation of Gold(II) Complexes. A Density Functional Study of Ligand and Solvent Effects," J. Phys. Chem. B 2006, 110(30), 14645-14651 (https://doi.org/10.1021/jp062501y).