As described above, the first player loses if $N$, the initial number of marbles, is a Fibonacci number, and wins for any other integer $N > 1$. Proof follows.
First, a definition.
The remainder of any integer $k$ is the smallest entry in a list of distinct non-consecutive Fibonacci numbers that add up to $k$. (For Fibonacci numbers, the remainder of $k$ is $k$.)
Proof of uniqueness: Find the largest integer $i$ such that $F(i) \leq k$. Start the list with $F(i)$. Then iterate by finding the list for $k - F(i)$. Note that at each step, you must be skipping a Fibonacci number, since $F(i) + F(i-1) = F(i+1)$. Thus, if $F(i)$ and $F(i-1)$ would be in the list, then both numbers should be excluded and $F(i+1)$ included, instead.
This process provides a deterministic formula for finding the list of non-consecutive Fibonacci numbers that add up to $k$. Since the ratios of non-consecutive Fibonacci numbers is always $\geq 2$, finding a second list that adds up to the same $k$ would be equivalent to (harder than) finding two distinct binary expressions for the same positive integer, which is impossible.
Strategy: If you see $N$ marbles in front of you, determine the unique list of non-consecutive Fibonacci numbers that add up to $N$, the smallest of which is the remainder of $N$. If you are allowed to remove the remainder number of marbles, do so, and you will win. If you cannot, you will lose, no matter what you do.
This strategy works because if you make the list shorter on your turn by removing the remainder of the current number of marbles, your opponent will not be able to make the list shorter on their turn, and if you do not make the list shorter on your turn, your opponent will be able to make the list shorter on their turn.
Proof: If you remove the remainder, $F(i)$, then you make the list shorter. Either you win immediately (because there was only one entry in the list), or the remainder of the new number of marbles at least $F(i+2) = F(i) + F(i+1) > 2*F(i)$. (Detail: while $F(3) = 2 = 2*F(1)$, you could not have both those entries in the list, as they would be replaced with $F(4) = 3$). Thus, if you make the list shorter, your opponent will never be able to make the list shorter on their next turn.
If, however, you cannot shorten the list, then you must remove some number of marbles $k$ less than the remainder of $N$. Let $F(i)$ be the remainder of $k$. The remainder of $N - k$ will always be either $F(i+1)$ or $F(i-1)$. This is because if $a+b=c$ (all positive integers), then the only way the remainder of $c$ can be greater than the remainders of both $a$ and $b$ is if the remainders of $a$ and $b$ are consecutive Fibonacci numbers. Otherwise, when you combine the sequences, the smaller remainder will always be left behind.
Note, also, that the ratio of two Fibonacci numbers is always $\leq 2$. Thus, if you remove $k$ marbles from $N$, where $k$ is less than the remainder of $N$, then the remainder of $N-k$ will be the next Fibonacci number higher or lower than the remainder of $k$. Since the remainder of $k$ cannot be greater than $k$, the remainder of $N-k$ must be less than or equal to $2k$. Thus, your opponent can always remove the remainder of $N-k$, shortening the list.
Thus, the only time you cannot win is if you cannot remove the remainder of the initial number of marbles $N$. This only occurs when $N$ is a Fibonacci number. For any other positive integer, you have a winning strategy.