I think that the answer here might be that you're comparing apples and oranges.
Let $F(x)$ denote the cdf of the Mann-Whitney $U$ statistic. qwilcox
is the quantile function $Q(\alpha)$ of $U$. By definition, it is therefore
$$Q(\alpha)=\inf \{x\in \mathbb{N}: F(x)\geq \alpha\},\qquad \alpha\in(0,1).$$
Because $U$ is discrete, there is usually no $x$ such that $F(x)=\alpha$, so typically $F(Q(\alpha))>\alpha$.
Now, consider the critical value $C(\alpha)$ for the test. In this case, you want $F(C(\alpha))\leq \alpha$, since you otherwise will have a test with a type I error rate that is larger than the nominal one. This is usually considered to be undesirable; conservative tests tend to be prefered. Hence,
$$C(\alpha)=\sup \{x\in \mathbb{N}: F(x)\leq \alpha\},\qquad \alpha\in(0,1).$$
Unless there is an $x$ such that $F(x)=\alpha$, we therefore have $C(\alpha)=Q(\alpha)-1$.
The reason for the discrepancy is that qwilcox
has been designed to compute quantiles and not critical values!