First note that we can think of the row as a row of $0$'s and $1$'s because we only care about pairity.
Let $n$ be the length of the row.
Claim:
$1)$. if $n$ is even and $A$ begins, than $A$ wins.
$2)$. if $n$ is odd and $B$ begins, than $A$ wins.
Note that $1)$ is enough for us for the problem, but we will prove both $1$ and $2$ simultaneously by induction on $n$.
The base cases $n=1,2,3$ are trivial. Suppose now the claim holds for $1,2,...,n-1$ and let's prove it for $n$.
If $n$ is even, the row is of the form $1,0,1,0,1,0,....,1,0$ and we may take the last pair of $(1,0)$ and make it a $1$. By induction hypothesis, since now it is $B$'s turn, $A$ wins.
If $n$ is odd and $B$ begins, the row is of the form $1,0,1,0,1,0,....,1,0,1$.
Player $B$ might pick a pair of the form $(1,0)$ or a pair of the form $(0,1)$, and replace this pair with a $0$ or a $1$. Note that doesn't matter what pair he chose to replace, there is either a $1$ right before it or a $1$ right after it: there is a $1$ after each pair of the form $(1,0)$ and there is a $1$ before each pair of the form $(0,1)$. Therefore the two numbers $B$ chose to replace are part of $3$ consecutive numbers in the row, which are $1,0,1$ (and $B$ chose either the first two or the last two of those three). Doesn't matter what $B$ did, we can make the $1,0,1$ that $B$ chose his pair from a $1$. For example, if $B$ chose the first two and made them a $0$, the $1,0,1$ became a $0,1$, which $A$ can make a $1$ from. Therefore, after $A$'s second turn, the difference from the original row is that a $1,0,1$ was replaced by a $1$. Now, since it is $B$'s turn now, by induction hypothesis $A$ wins.
Your question is a special case of this one, and the answer there proves that $A$ wins on all connected regions with an odd number of squares.
A winning strategy is to consider a partial domino tiling of the region which has the most dominos out of all such partial tilings. $A$'s first move will be to remove any square not covered by a domino. Whenever $B$ removes one square of a domino, $A$ responds by removing the other. You can show that the maximality of the partial domino tiling implies that $B$ can never remove a square which is not covered by a domino, so $A$ can always apply this strategy, and is never without a move.
The same proof applies to any region with an even number of squares which cannot be completely tiled by dominos. If the region can be completely tiled by dominos, then $B$ wins with a similar strategy.
Best Answer
If the moves consist of taking either $a$ or $b$ sticks, with $a\lt b$ and, as clarified in a comment, $N\lt a$ is a loss, then the first player wins iff $((N-1)\bmod(a+b))\bmod(2a)\lt a$, and a winning strategy is to take $a$ sticks if $(N-1)\bmod(a+b)\gt b$, take $b$ sticks if $(N-1)\bmod(a+b)\lt a$, and take either otherwise.