As a general rule, you'll want to use Zeckendorf's theorem to find these representations; it says that any number has a unique representation as the sum of nonconsecutive Fibonacci numbers. As André has noted, your conjecture is false in general, but it is possible to find a straightforward bound on the number of possible sums.
Let's say you want to write $k$ as the sum of $n$ Fibonacci numbers, and the Zeckendorf's theorem sum contains $m terms. From any $n$-term sum you like, you'll be able to get the Zeckendorf sum by repeatedly combining terms until no remaining terms are consecutive. This implies that you can get any $n$-term sum by starting with the Zeckendorf sum and splitting up terms in it until you get $n$ of them.
Say you start with a single term $F_a$. The only way to split that is as $F_a=F_{a-1}+F_{a-2}$; the only way to split that is as $F_{a-1}+F_{a-3}+F_{a-4}$; and so on -- that is, any sum that began as a single Fibonacci number has at most one representation as the sum of $n$ distinct Fibonacci numbers for any $n$ (since at each point the indices differ by at most two, the only additional split that's possible comes from breaking up the smallest number in the sum).
So, if you start with a Zeckendorf sum that has $m$ terms, your only possible decision is how many times to split up each of those terms; once you know that, you'll have a unique sum. (It may be that your terms will "collide" and you won't get a sum of distinct Fibonacci numbers, but you'll certainly get every possible sum from some such specification, so this leads to an overcount.) To get an $n$-term sum, you need to split a total of $n-m$ times. By a stars-and-bars argument, there are at most ${n-1 \choose m-1}$ ways of doing this. So a repaired version of your conjecture would state:
If the Zeckendorf sum for $k$ contains $m$ terms, there are at most ${n-1 \choose m-1}$ ways of writing $k$ as the sum of $n$ distinct Fibonacci numbers.