It will likely not work with other halides. As far as I understood the winkle method, the oxygen is first fixated with $\ce{Mn^2+}$ in basic conditions:
$$\begin{align}
\ce{Mn^2+ (aq) + 2 OH- (aq)&-> Mn(OH)2 (aq)}\tag{1}\label{alkalinemanganese}\\
\ce{2Mn(OH)2 (aq) + O2 (aq) &-> 2 MnO(OH)2 (s) v}\tag{2}\label{bindingoxygen}\\
\end{align}$$
The sample is acidified and iodide is added. This is oxidised to $\ce{I2}$, which forms a complex with remaining iodide.
$$\begin{align}
\ce{MnO(OH)2 (s) + H+ (aq) &-> Mn(OH)3 (aq)}\tag{3}\label{acidicmanganese}\\
\ce{2 Mn(OH)3 (aq) + 2 I- (aq) + 6 H+ (aq)
&-> 2 Mn^2+ (aq) + I2 (aq) + 6 H2O}\tag4\label{iodide-ox}\\
\ce{I2 (aq) + I- (aq) &<=> I3- (aq)}\tag5\label{triiodide}\\
\end{align}$$
The $\ce{I3-}$ forms the indication complex with the starch, which can then be titrated with thiosulfate solution. When the colour is gone you have successfully converted all of the prepared $\ce{MnO(OH)2}$ back to manganese(II) ions.
$$\ce{I3- (aq) + 2 S2O3^2- (aq) -> 3 I- (aq) + S4O6^2- (aq)}\tag6\label{triiodide-reduction}$$
The reason why this works is because you have the equilibrium $\eqref{triiodide}$ which produces the triiodide ion, which is relatively stable in aqueous solution. Iod by itself does not dissolve in water (much). This complex again forms a very colourful complex with starch, the switch can easily be detected by the human eye.
Polyatomic ions of other halogens are not unheard of (ref. Trichloride ion (c.f. triiodide ion) and therein), but they are a lot less stable. I am not aware that they form a complex with starch so they won't be that easily detected.
Another aspect, I don't want to go much deeper here, is that the redox reactions necessary $(\ref{iodide-ox},\ref{triiodide-reduction})$ for this procedure to work might not work at all with other halides.
The original method was proposed in the 19th century and is still applied today. It is a quite accurate and robust method. There is not really a reason to tamper with it.
