Indeed, the hunter has a winning strategy, for example, the one given below. First, he identifies the grid with the set $\mathbb{Z}^{2}$, then associates each cell $\mathbf{c}=(x, y)$ with the triplet $\varphi(\mathbf{c})=(x, y, x+y)$.
Each time the rabbit moves, two components of the triplet $\varphi \mathbf{( c )}$ vary by $\pm 1$, and the third component does not change. Furthermore, knowing two of the coordinates of $\varphi(\mathbf{c})$ is enough to identify c. In particular, if there exists a finite set $E$ such that at any time, the rabbit is in a cell c such that at least two of the components of $\varphi(\mathbf{c})$ are in $E$, it is known that the rabbit is constrained to evolve within a finite set of cells, which will inevitably lead to its defeat and transformation into civet.
Therefore, at least two of the three components of $\varphi(\mathbf{c})$ will take an infinite number of values. If the hunter can identify, at a given moment, the values taken by these components, he will deduce the actual position of the rabbit and will have won.
By doing so, the hunter has reduced the problem to the following one-dimensional problem: given an invisible rabbit that moves along the axis $\mathbb{Z}$ by steps of $-1,0$ or +1 and is constrained to pass through an infinite number of integers, how can coloring $\mathbb{Z}$ using a finite number of colors allow the hunter to identify the initial position of the rabbit along this axis? Once such a coloring col: $\mathbb{Z} \rightarrow\{0,1, \ldots, \kappa-1\}$ using $\kappa$ colors numbered from 0 to $\kappa-1$ is established, the hunter only needs to use $\kappa^{3}$ colors, identified with the elements of the set $\{0,1, \ldots, \kappa-1\}^{3}$, and to color the cell $\mathbf{c}=(x, y)$ with the color $(\operatorname{col}(x), \operatorname{col}(y), \operatorname{col}(x+y))$.
To answer this new one-dimensional problem, the hunter colors $\mathbb{Z}$ using three colors, numbered from 1 to 3, and paints in color $c$ all integers $n$ such that $n \equiv c(\bmod 3)$. In this way, he knows at any time whether the rabbit is moving to the left or to the right.
Then he repaints with color 4 the integers, previously colored 3, that are of the form $3 \times(-3)^{n}$ with $n$ a natural number. Thus, he does not lose his previous information on the value of the cells modulo 3, nor on the value $(-1,0$ or +1) of the rabbit's movements. Consequently, if he ever manages to identify the rabbit's position at a given moment, he knows its position at any time during the game.
Finally, the rabbit will pass through two consecutive elements of the set $\mathcal{E}=\left\{3 \times(-3)^{n}: n \geqslant 0\right\}$. If these two elements are $3 \times(-3)^{n}$ and $3 \times(-3)^{n+2}$, the difference between them, in absolute value, is $8 \times 3^{n}$. If they are 3 and -9, this difference, in absolute value, is equal to 12. Therefore, after the rabbit has passed through two consecutive elements of $\mathcal{E}$, the hunter easily identifies these two elements and, based on the variations in the rabbit's position, the element of $\mathcal{E}$ where the rabbit has just arrived. He then deduces the initial position of the rabbit, which allows him to win the one-dimensional game, and thus win the original two-dimensional game.