I'm going to prove that no player can guarantee the win. First, among $31, 32, \dots, 39, 71, 72, \dots, 79$, only $36, 76$ are quadratic residues $\mod100$ as quadratic residues must $\equiv0, 1\pmod4$, and $\equiv0, \pm1\pmod5$. Second, if two squares are differ by $40$, then they must be $9, 49$ or $81, 121$, which can be found by solving $(a+b)(a-b)=40$. Consider the strategy that in a player's round, it chooses $3$ or $7$ to write at the end such that writing an additional $6$ at the end won't make the number a perfect square. From the above, we know that no two squares that end with $6$ are differ by $40$, so the strategy is legal. In the next round, it is impossible to make the number a square as no square ends with $3$ or $7$, and adding any digit to the end won't making it a square by the strategy. Also, $76$ is not a perfect square. $\therefore$ under this strategy, no player can win.

No player guarantees to win. A perfect square cannot end with $2,3$ since they are not quadratic residues in $mod \ 5$. First player cannot obtain a perfect square in first move. The difference between two $\geq 3$ digit consecutive perfect squares is at least $21$. Thus at least one of $\overline{a_n...a_02k}$ and $\overline{a_n...a_03k}$ is not perfect square. Players can write $2$ or $3$ where a perfect square won't be able to obtained in next turn since a perfect square cannot end with $2,3$ and $\overline{a_n...a_0\{2,3\}k}$ cannot be perfect square for $k=0,...,9$.$\blacksquare$