As Ross pointed out, you can solve this using nimbers. Working out the nimbers for some low coin counts leads to the winning strategy: Always make sure the coin counts are roughly equal, $\pm1$, and that their sum is divisible by $3$. That is, if the coin counts are equal and both have remainder $1$ mod $3$, take one coin from each pile; if the coin counts are equal and both have remainder $2$ mod $3$, take a coin from either pile: if the coin counts are different, take as many from the larger pile that you end up with the same count $\pm1$ and the sum is divisible by $3$. Such a move is always possible if the position isn't one of the losing positions $(3k,3k)$ or $(3k+1,3k+2)$ or $(3k+2,3k+1)$.
In particular, since $2010$ is divisible by $3$, the initial position is a losing position for player $A$.
Background and Notation
A single Nim heap of size $n$ will be denoted "$*n$", with $*1$ abbreviated "$*$" and $*0$ abbreviated "$0$".
"$+$" denotes the disjunctive sum, so that $*+*3$ is a Nim position with one heap of size $1$ and one heap of size $3$.
This is not standard, but we can take a page from Topology and call a game $G$ connected if it is not isomorphic to a disjunctive sum other than $G+0$ and $0+G$.
Edit: "$\vee$" denotes a variant of the selective sum, so that when $G$ and $H$ are nim heaps, $G\vee H$ is a game in which a player can move in $G$ or $H$ or both simultaneously but if both then only one token/chip is removed from each.
The game described by user89 is one in which there are two Nim heaps (one being $*$), and a player can move in one or both of them on their turn. This means that this game with $n+1$ chips is $*n\vee*$.
Question:
Is $*n\vee*$ connected for each $n$?
Nim heaps are connected
$0$ and $*$ are connected since a nontrivial disjunctive sum must have a line of play (a "run") of length at least $2$, since each summand must have at least one move.
For $n\ge2$, $*n$ is connected since it has a move to $0$, but a disjunctive sum $G+H$ does not if neither $G$ nor $H$ are $0$.
The OP's game is connected
Note that $0\vee*$ is isomorphic to $*$, which is connected.
Note that $*\vee*$ is isomorphic to $*2$, which is connected.
If $n\ge2$, then $*n\vee*$ has a move to $0\vee*$ and a move to $*\vee*$. Suppose $*n\vee*=G+H$ with $G,H$ nonzero. Since there is a move to $*$, then one of $G$ and $H$ must be $*$ since otherwise there is no run of length $2$. Suppose $H$ is $*$. But, then, since there is a move to $*2$, and $*2$ is connected, we $G$ must be $*2$. For $n>2$ we have a contradiction since $*n\vee*$ has a longer run than any run of $*2+*$.
For $n=2$, we have a contradiction since $*2+*$ has a move to $*+*$, but $*2\vee*$ only has moves to $*2$ (two of them) and to $*$.
A generalization
Suppose, for sake of induction, that for some $k\ge 1$ we have shown that $*n\vee*k$ is connected. Then if we have a position like $*m\vee*(k+1)$, it has moves to $*(k+1)$ and to $*m\vee*k$, both of which are connected, distinct (unless $m=1$ and $k=1$, which was handled earlier) and nonzero. Then if $*m\vee*(k+1)$ is a disjunctive sum, it must be the sum of those two. But the length of the longest run in the disjunctive sum is $m+2k+1$, and the length of the longest run in $*m\vee*(k+1)$ is the smaller $m+k+1$.
Best Answer
You had the right idea filling out a 10 by 10 table. Everything you said is a winning position is indeed winning. However, you did not list all of the winning positions in the $8\times 7$ table, so you either made mistakes, or have not shown all your work. I cannot tell what you have done so far, so I will start from the ground up.
Recall the rules:
If all of a position's options are winning, or there are no options, then that position is losing. This is because you either have no moves, or your are forced to hand your opponent a winning position. A losing position is also called a $P$-position, meaning that it favors Previous player.
If a position has at least one losing option, then it is winning. The winning strategy is to select that losing option, which forces your opponent to lose. A winning position is also called an $N$-position, meaning it favors the Next player.
Let me explain. You were correct that $(1,x)$ is winning when $x\ge2$. Furthermore, $(1,1)$ is losing, as there are no options.
Next, look at (2,2). Both of its options, (1,0) and (0,1), are losing. Therefore, (2,2) is winning.
Similarly, $(2,x)$ is winning when $x\ge 3$. The winning move is to $(0,x-1)$.
Next, we look at $(3,3)$. Its options are $(1,2)$ and $(2,1)$, which we previously determined are losing.
Similarly, you can see that $(3,x)$ is losing when $x\ge 4$, because both $(1,x-1)$ and $(2,x-2)$ were both determined to be winning.
Next, we look at $(4,4)$. The two options are $(2,3)$ and $(3,2)$. Both of these are winning, so $(4,4)$ is losing.
Next, we look at $(4,5)$. This has a winning move to $(3,3)$, because $(3,3)$ is losing, so $(4,5)$ is winning. Similarly, you can show that $(4,x)$ is winning for all $x\ge 5$.
Here is the correctly filled out table, so far. I have also included the cases where one of the piles is zero, as these arise in the game.
$$ \begin{array}{c|cccccccccc} & 0&1&2&3&4&5&6&7&8&9&10 \\\hline 0 & P&P&P&P&P&P&P&P&P&P&P& \\ 1 & P&P&N&N&N&N&N&N&N&N&N& \\ 2 & P&N&N&N&N&N&N&N&N&N&N& \\ 3 & P&N&N&P&P&P&P&P&P&P&P& \\ 4 & P&N&N&P&P&N&N&N&N&N&N& \\ 5 & P&N&N&P&N \\ 6 & P&N&N&P&N \\ 7 & P&N&N&P&N \\ 8 & P&N&N&P&N \\ 9 & P&N&N&P&N \\ 10 & P&N&N&P&N \end{array} $$
Now, here is the task for you:
Fill out the rest of this table. Remember, for each entry, you need to check its two options. For the entry $(x,y)$, you need to check $(x-1,y-2)$ and $(x-2,y-1)$. If these are both winning (both $N$), then $(x,y)$ is losing ($P$). If at least one is losing (at least one $P$), then $(x,y)$ is winning ($N$).
Find a pattern in the table entries. Describe it in words, or mathematically.
Prove that the pattern holds forever, by giving a winning strategy for all $(x,y)$ which the pattern says are $N$. This is the hard part. You need to find some "property" that all of the $P$-positions have, but none of the $N$-positions do. You then need to prove two things:
Every $N$-position has a $P$ option. That is, for every posiiton without the "property", one of its options does have the "property".
Every $P$ position has no $P$ options. That is, for every position with the property, none of its options have the "property".
Solution
I encourage you to attempt to solve the question without reading below, and then use below to check your work, or if you are horribly stuck.