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Let $c=⟨R_0|R_1⟩$. The joint state after the environment interaction is $\alpha |0⟩|R_0⟩+\beta |1⟩|R_1⟩$. Taking its partial trace over the environment gives the exact reduced density operatorThe conjugate in the upper off-diagonal entry follows from tracing $|R_0⟩⟨R_1|$. Normalization of the environment states and the Cauchy-Schwarz inequality imply $|c|\le 1$.

For comparison, a phase-flip channel sends $\rho$ to $(1-p)\rho +pZ\rho Z$, multiplying its off-diagonal entries by the real number $1-2p$. Consequently the claimed representation is valid, for a general input, provided the environment overlap is real, withThis is a genuine probability between zero and one. The environment maps printed in the question do not themselves imply that $c$ is real. For example, choose $|R_1⟩=i|R_0⟩$, so $c=i$, and take $\alpha =\beta =1/\sqrt{2}$. The qubit becomes $(|0⟩+i|1⟩)/\sqrt{2}$, whose imaginary off-diagonal entries cannot come from any probabilistic mixture of $\rho$ and $Z\rho Z$. These environment maps are a valid isometry and can be extended to a unitary interaction, so this is an actual counterexample to the unrestricted premise.

The general complex environment overlap in qubit dephasing has an exact corrected description. Write $c=r{e}^{i\phi }$, $0\le r\le 1$, and let $D_{\phi }=\mathrm{diag}(1,e^{i\phi })$. ThenAt $r=0$, choose any phase. The dephasing channel includes both random phase errors and, when the overlap is complex, a coherent diagonal rotation. The subsequent probability formulas use the intended real-overlap phase-flip model.