Let the square be $[0, 6] \times [0, 6]$. We first show that either some grid-vertical $x = i$, $i = 1, 2, 3, 4, 5$, or some grid-horizontal $y = j$, $j = 1, 2, 3, 4, 5$, crosses no tile. Let $m_i$ and $n_j$ be the number of tiles crossed by the grid-vertical $x = i$ and the grid-horizontal $y = j$, respectively. Clearly, a grid-vertical crosses only horizontal tiles, and a grid-horizontal crosses only vertical tiles.

Since every horizontal tile is crossed by a single grid-vertical, the number of horizontal tiles is $m_1 + m_2 + m_3 + m_4 + m_5$. Similarly, the number of vertical tiles is $n_1 + n_2 + n_3 + n_4 + n_5$. Hence the number of tiles is $m_1 + m_2 + m_3 + m_4 + m_5 + n_1 + n_2 + n_3 + n_4 + n_5 = 18$. Consequently, either some $m_i \le 1$, $i = 1, 2, 3, 4, 5$, or some $n_j \le 1$, $j = 1, 2, 3, 4, 5$.

Now, for each positive integer $i \le 5$, the rectangle $[0, i] \times [0, 6]$ consists of a certain number of tiles and $m_i$ unit cells, the left halves of the horizontal tiles the grid-vertical $x = i$ crosses. Since the area of each $[0, i] \times [0, 6]$ and the area of each tile are both even, so is each $m_i$. Similarly, each $n_j$ is even.

Finally, by the conclusion of the preceding paragraph, either some $m_i = 0$, $i = 1, 2, 3, 4, 5$, in which case the corresponding grid-vertical $x = i$ crosses no tile; or some $n_j = 0$, $j = 1, 2, 3, 4, 5$, in which case the corresponding grid-horizontal $y = j$ crosses no tile.

Alternative solution.

Suppose, in the above setting, that every grid-line, whether vertical or horizontal, crosses at least one tile. Then the five $m_i$ and the five $n_j$ are all positive even integers, i.e., they are all at least $2$. Consequently, the ten add up to at least $20 > 18$ which is a contradiction. This establishes the first part.

The answer to the second part is in the affirmative. To prove this, we exhibit a domino tiling of the square $[0, 6] \times [0, 6]$ with a single separating line, i.e., one crossing no tile. Clearly, grid-lines alone are to be considered.

Begin by tiling the rectangle $[0, 3] \times [0, 6]$ by four horizontal dominoes, namely,
$$[0, 2] \times [0, 1], \quad [0, 2] \times [3, 4], \quad [1, 3] \times [4, 5], \quad [1, 3] \times [5, 6],$$
and five vertical dominoes, namely,
$$[2, 3] \times [0, 2], \quad [0, 1] \times [1, 3], \quad [1, 2] \times [1, 3], \quad [2, 3] \times [2, 4], \quad [0, 1] \times [4, 6].$$

Notice that the grid–horizontal $y = 4$ is the single separating line of this tiling.

Next, tile the rectangle $[3, 6] \times [0, 6]$ by a copy of the reflection of the above tiling in the grid–horizontal $y = 3$, to make the grid–horizontal $y = 2$ its single separating line. Explicitly, the four horizontal tiles are
$$[4, 6] \times [0, 1], \quad [4, 6] \times [1, 2], \quad [3, 5] \times [2, 3], \quad [3, 5] \times [5, 6],$$
and the five vertical tiles are
$$[3, 4] \times [0, 2], \quad [5, 6] \times [2, 4], \quad [3, 4] \times [3, 5], \quad [4, 5] \times [3, 5], \quad [5, 6] \times [4, 6].$$
The grid–horizontal $y = 2$ is clearly the single separating line of this tiling.

Finally, the two tilings fit together along the grid–vertical $x = 3$ to form an overall tiling of the square $[0, 6] \times [0, 6]$ with a single separating line — the grid–vertical $x = 3$, of course.