"Generally Octahedral" - Meaning?
The term "generally octahedral" refers to a distorted octahedral arrangement of the ligands referring to the central atom (as ron already stated). In this case the distortion is due to the massively different bond lengths.

In addition to ron's answer I did a little calculation at BP86/cc-pVDZ. I did not use any symmetry constraints at first and the final geometry deviated from ideal $C_\mathrm{4v}$ geometry only slightly due to rounding errors. After symmetry constraint reoptimisation, the following parameters may be found:  

\begin{array}{lrl}
\text{parameter}                &amp; \text{value}&amp;\\\hline
\mathbf{d}(\ce{S-Br})           &amp; 2.039 &amp;\mathrm{\mathring{A}}\\
\mathbf{d}(\ce{S-F_{ax}})       &amp; 1.638 &amp;\mathrm{\mathring{A}}\\
\mathbf{d}(\ce{S-F_{eq}})       &amp; 1.650 &amp;\mathrm{\mathring{A}}\\\hline
\angle(\ce{F_{ax}-S-Br})        &amp; 180.0 &amp;{}^{\circ}\\
\angle(\ce{F_{eq}-S-Br})        &amp;  90.7 &amp;{}^{\circ}\\
\angle(\ce{F_{ax}-S-F_{eq}})    &amp;  89.3 &amp;{}^{\circ}\\
\angle(\ce{F_{eq}-S-F'_{eq}})   &amp;  90.0 &amp;{}^{\circ}\\
\angle(\ce{F_{eq}-S-F''_{eq}})  &amp; 178.7 &amp;{}^{\circ}\\\hline
\angle_\mathrm{D}(\ce{F_{eq}-S-F'_{eq}-F''_{eq}})  
                                &amp; 178.7 &amp;{}^{\circ}\\\hline
\end{array}



I was unable to determine a interconversion barrier related to the $\angle(\ce{F_{eq}-S-Br})$ bond angle change. I was also not able to find a different stable isomer.

In solid state it is most likely, that the geometric parameters will change due to packing effects and/or more internal entropy.



Addressing the edit:


  Are you sure there are 8 F-S-F bond angles &lt; 90 degrees? I'm can see how the Br pushes down on the fluorines in the equatorial plane, creating four bond angles &lt; 90 degrees. Where's the other four? Or are these from taking the Br to be in the equatorial plane?


Yes, the bromine pushes down the fluorines of the equatorial plane resulting in $\angle(\ce{F_{ax}-S-F_{eq}})=89.3^{\circ}$ bond angle. Due to this, the sulfur is out of plane with the equatorial fluorine atoms. Since the planes expanded by $\ce{Br-S-F_{eq}}$ and $\ce{Br-S-F'_{eq}}$ have to be at $90^\circ$ due to symmetry constraints, this is creating also a slightly smaller angle than $90^\circ$ for the equatorial fluorines. In this very case it can be found to be $\angle(\ce{F_{eq}-S-F'_{eq}})=89.992^{\circ}$, which was rounded by the program to fit the one decimal place restriction. The application of this is just trigonometry and you can derive it as a simple thought experiment - if you like.