No, $d$ need not be different from both $b$ and $c$. For instance, if $A = \\{a,b,c\\}$, the rewrite system $(A, \\{ a \rightarrow b,\, a \rightarrow c,\,b \rightarrow c\\})$ is confluent, but the only way to join $b$ and $c$ is $b \rightarrow^1 d \leftarrow^0 c$ with $d$ identical to $c$. Remember that $x \rightarrow^* y$ means that there exists some $n \geq 0$ (!) with $x \rightarrow^n y$, so the case $n = 0$ is included.