Does fluorine in FNO3 have +1 oxidation number?
tl;dr The ambiguity is due to an unfortunate incompleteness of the previously (prior to 2016) existing rules. The oxidation state of fluorine in $\ce{FNO3}$ is $-1$ according to the present rules.

As a general reminder, it is important to understand, that oxidation states are a bookkeeping tool only, and that they hardly represent the general bonding in molecules. With that being said, it is a very simple and powerful tool in many situations. As it is very simple, one has to especially pay attention to its shortcomings or ambiguities.

Prior to 2016 the IUPAC recommended an incomplete set of rules, which I discussed in more detail in Electronegativity Considerations in Assigning Oxidation States.

The special case you have cited is very close to the case of $\ce{HOF}$, which has been previously discussed by me on two occasions already: Oxidation state of fluorine in HOF, What should be the oxidation state of oxygen in HOF (hypofluorous acid)?

According to the new rules, which are based on Allen electronegativities (see Pure Appl. Chem. 2016, 88 (8), 831–839 for more detail), and of which the sort version is below adapted from the gold book, there is no more ambiguity about fluorine, which will be assigned $\text{OS}(\ce{F}) = -1$.


  oxidation state
  gives the degree of oxidation of an atom in terms of counting electrons. The higher the oxidation state (OS) of a given atom, the greater is its degree of oxidation. Definition:
  OS of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds.


Applied to $\ce{FNO3}$, you would assign fluoride an oxidation state of $\text{OS}(\ce{F})=-1$, the terminal oxygens $\text{OS}(\ce{O})=-2$, the nitrogen will have its highest possible state $\text{OS}(\ce{N})=+5$, and the bridging oxygen will have $\text{OS}(\ce{O})=0$. 



Contrary to that are the pre-2016 rules by the IUPAC according to the gold book (via the Internet Archive):


  A measure of the degree of oxidation of an atom in a substance. It is
  defined as the charge an atom might be imagined to have when electrons
  are counted according to an agreed-upon set of rules: 
  
  
  the oxidation state of a free element (uncombined element) is zero; 
  for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion;
  hydrogen has an oxidation state of 1 and oxygen has an oxidation state of -2 when they are present in most compounds. (Exceptions to
  this are that hydrogen has an oxidation state of -1 in hydrides of
  active metals, e.g. $\ce{LiH}$, and oxygen has an oxidation state of
  -1 in peroxides, e.g. $\ce{H2O2}$;
  the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the
  oxidation states of the constituent atoms must be equal to the charge
  on the ion. For example, the oxidation states of sulfur in $\ce{H2S}$,
  $\ce{S8}$ (elementary sulfur), $\ce{SO2}$, $\ce{SO3}$, and
  $\ce{H2SO4}$ are, respectively: -2, 0, +4, +6 and +6. The higher the
  oxidation state of a given atom, the greater is its degree of
  oxidation; the lower the oxidation state, the greater is its degree of
  reduction.
  


You can already see that there is a problem in these rules since they are not definitive. The keyword here is most in bullet 3. They are at least incomplete since there is no mentioning of other elements than oxygen and hydrogen.

This will obviously lead to contradictory results. According to the deprecated rules we need to give oxygen an oxidation state of $\text{OS}(\ce{O})=-2$ in $\ce{NO3F}$. This will give us a total of $6$ electrons that need to come from the other elements. Since nitrogen only has five valence electrons, anything larger than $\text{OS}(\ce{N})=+5$ does not make any sense. Unfortunately that leaves one electron, which can only come from fluoride, i.e. $\text{OS}(\ce{F})=\color{red}{+1}$. It is obviously ridiculous that the most electronegative element has a higher oxidation state than one with a lower electronegativity. This is also in no way an accurate representation of the charge distribution in the molecule.

 

The above image shows the (NPA) charges of the molecule calculated on the DF-BP86/def2-SVP level of theory. It clearly shows, that fluoride is not positively charged.

One really has to understand that an oxidation state of $+1$ for fuorine does not make any physical sense, and that this set of rules is insufficient. Thankfully this has been addressed in the new definition, which is not perfect, but at least complete.