Answer: Bob wins.
It is well-known that a natural number $n$ is divisible by $11$ if and only if $(S_o - S_e) \div 11$, where $S_o$, $S_e$ are the sums of the digits on the odd and even positions respectively in the decimal representation of $n$.
Let $2n + 1$ ($n \in \mathbb{N}$) asterisks be written on the blackboard: $\underbrace{**\dots*}_{2n+1}$.
Show Bob's winning strategy. If Ann replaces some asterisk (different from the first one) by $c$, then, in answer, Bob replaces asterisk of the opposite parity by the same digit $c$ (note that Bob always chooses the asterisk different from the first one). Thus, if at the end Ann replaces the first asterisk by some digit $d$, then $S_o - S_e = d$. Since $d \neq 0$ we see that the obtained number is not divisible by $11$.
If Ann replaces the first asterisk by some digit $c \neq 0$ and it is not her last move, then Bob replaces some even asterisk by $c-1$, and further he keeps the strategy described above. As the result in the end Ann must replace some odd asterisk by some digit $d$. Therefore, in this case $S_o - S_e = d + c - (c - 1) = d + 1$. Since $d$ is a digit we have $0 < d + 1 < 11$, so the obtained number is not divisible by $11$, and Bob wins.