= Solution
Assume a finite-dimensional memoryless <quantum channel>, no shared <entanglement>, and a code of $M_n$ messages with $n^{-1}\log_2M_n\to R$. Use a uniform message $M$ to relate average and maximum error. Each permitted input is a <product state> $\rho_m^{(n)}=\bigotimes_{i=1}^n\rho_{m,i}$, giving output $\sigma_m^{(n)}=\bigotimes_i\Lambda(\rho_{m,i})$. The factors may depend on both the message and the channel use.
Let $\chi^*(\Lambda)$ denote the one-use <Holevo capacity> in the question. The <Subadditivity of Von Neumann entropy> for the average output, and entropy additivity for each product output, imply
$$
\begin{aligned}
\chi_n
&=S\left(\frac1{M_n}\sum_m\sigma_m^{(n)}\right)
-\frac1{M_n}\sum_m\sum_iS(\Lambda(\rho_{m,i}))\\
&\leq\sum_i\left[S\left(\frac1{M_n}\sum_m\Lambda(\rho_{m,i})\right)
-\frac1{M_n}\sum_mS(\Lambda(\rho_{m,i}))\right]
\leq n\chi^*(\Lambda).
\end{aligned}
$$
This is the key step of the <product-input classical-capacity converse>; it does not assume that the average output itself is a product.
Bob may measure all outputs jointly. The <Holevo bound> still gives $I(M:\widehat M)\leq\chi_n$. Let $P_e$ be his average message-error probability. <Fano's inequality> gives
$$
\log_2M_n=I(M:\widehat M)+H(M\mid\widehat M)
\leq n\chi^*(\Lambda)+1+P_e\log_2M_n.
$$
Since the maximum error $P_{\max}$ is at least $P_e$,
$$
\boxed{\liminf_{n\to\infty}P_{\max}
\geq1-\frac{\chi^*(\Lambda)}R>0}
$$
when $R>\chi^*(\Lambda)$. Thus the maximum error cannot vanish asymptotically.
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