We show that Quintijn has a winning strategy.
Observe that each move reduces the number of integers by exactly one, so at the start of Johan's turn the number is always even and at the start of Quintijn's turn the number is always odd. Moreover, the number must be at least $2$, otherwise the game would have already ended. In particular, at the start of Quintijn's turn, the number of integers is always at least $3$.

Note Johan cannot reach a position in which $x_1 = x_2 = 0$ unless at the start of his turn either $x_1 = 2$ and $x_2 = 0$, or $x_1 = 0$ and $x_2 = 2$ (or $x_1 = 0$ and $x_2 = 0$) hold. In particular, if at the start of his turn at least one of $x_1$ and $x_2$ is odd, then he cannot reach a position in which $x_1 = x_2 = 0$.

First suppose that $x_1$ and $x_2$ are both even. Then at least one of $x_1$ and $x_2$ must be at least $2$, say $x_1 \ge 2$. Since Quintijn always has an odd number of integers left at the start of his turn, there must also be an integer on the board that is divisible by $3$. With $(0, 1) \to 2$, Quintijn ensures that $x_1$ decreases by $1$ and $x_2$ increases by $1$, so $x_1$ and $x_2$ both become odd.
Now suppose that at least one of $x_1$ and $x_2$ are odd. If there exists an $i$ with $x_i \ge 2$, then performing any move $(i, i) \to 0$ will not change the parity of $x_1$ and $x_2$, so at least one of them remains odd. Otherwise all $x_i \le 1$ so there are at most $3$ integers left, and in fact we must have equality here as there are always at least $3$ integers left at the start of Quintijn's turn. So $x_0 = x_1 = x_2 = 1$, hence Quintijn can perform the move $(1, 0) \to 1$.

Therefore Quintijn can always make a move that causes at least one of $x_1$ and $x_2$ to be odd. Now the strategy of making such a move is winning for Quintijn, as by following this strategy, Quintijn will never create a position in which $x_1$ and $x_2$ are both $0$, and Johan can never create such a position from any position that Quintijn may leave behind; the same must hold for the final position in which one integer remains, so this remaining integer must not be divisible by $3$, meaning that Quintijn wins. $\square$