I would say that the player going second has a better chance of winning. For example if there are only 6 sticks, like this
the second player always wins:
And vice versa (e.g. if Player 1 crosses D+E, Player 2 crosses B or C). The trick is to force your opponent into (i) an odd number of non-adjacent single sticks, or (ii) two non-adjacent rows of two adjacent sticks.
The game looks more complicated with two more rows added but I think the basic idea is the same. Note that (ii) can be generalized to: an even number of non-adjacent sets of n adjacent sticks.
PS: If there are 10 sticks (four rows), the first player will always wins – he just needs to cancel the whole of the last row and force the second player to make the first move in the 6-stick situation. So I think it’s because there are an odd number of rows that the second player is favoured.