Must $\vec{n}$ be a Unit Normal Vector (Stokes' Theorem)? If $S$ is an oriented, smooth surface that is bounded by a simple, closed, smooth boundary curve $C$ with positive orientation, then for some vector field $\vec{F}$:
$$\oint_C \vec{F} \cdot d\vec{r} =  \iint_S {\rm curl} \> \vec{F} \cdot d\vec{S}$$
The latter integral can be written equivalently as follows for some vector $\vec{n}$, given that it is normal to the surface $S$ and has the proper orientation:
$$\iint_S {\rm curl} \> \vec{F} \cdot \vec{n} \> dS$$
Ultimately, my question is whether or not this normal vector, $\vec{n}$, must be a unit normal vector or not (or if there are other constraints that must be imposed on it). The reason I ask this is because, while working on a problem involving Stokes' Theorem, I deduced that the appropriate normal vector for some surface was $\hat{i} + \hat{k}$, and so I normalized it, yielding $\frac{1}{\sqrt 2}\hat{i} + \frac{1}{\sqrt 2}\hat{k}$. 
My answer ended up being off by a factor of $\frac{1}{\sqrt 2}$ which makes me think that how I have defined $\vec{n}$ for these types of problems is incorrect.

[Edit] For those interested, the problem was to evaluate $\oint_C \vec{F} \cdot d\vec{r}$ for $\vec{F}(x, y, z) = xy\>\hat{i} + 2z\>\hat{j} + 6y\>\hat{k}$ such that $C$ is the counterclockwise-oriented curve of intersection of the plane $x + z = 1$ and the cylinder $x^2 + y^2 = 36$. 
 A: In fact, the only constraints for the vector $\bf{n}$ are
$1.$ The vector $\bf{n}$ is a unit vector normal to the surface.
$2.$ It should have proper orientation depending on the orientation of the surrounding curve.
So, I think you may have made a mistake in the problem you solved and hence we may help you if you write it down in your question. :)

Verifying Stokes Theorem For Your Question 
Your surface is enclosed by the intersection curve of the plane $x+z=1$ and the cylinder $x^2+y^2=36$ as the following figure shows. 

The parametric equation of the intersection curve, the tangent vector, and the vector field are
$$\eqalign{
  & {\bf{x}} = 6\cos \theta {\bf{i}} + 6\sin \theta {\bf{j}} + \left( {1 - 6\cos \theta } \right){\bf{k}}  \cr 
  & {{d{\bf{x}}} \over {d\theta }} =  - 6\sin \theta {\bf{i}} + 6\cos \theta {\bf{j}} + 6\sin \theta {\bf{k}}  \cr 
  & F({\bf{x}}) = xy{\bf{i}} + 2z{\bf{j}} + 6y{\bf{k}} \cr} $$
and hence the line integral will be
$$\eqalign{
  & I = \int\limits_C {F({\bf{x}}) \cdot {{d{\bf{x}}} \over {d\theta }}d\theta }  = \int_{\theta  = 0}^{2\pi } {\left( { - 6\sin \theta xy + 12\cos \theta z + 36\sin \theta y} \right)d\theta }   \cr 
  & \,\,\, = 6\int_{\theta  = 0}^{2\pi } {\left( { - 36{{\sin }^2}\theta \cos \theta  + 2\cos \theta \left( {1 - 6\cos \theta } \right) + 36{{\sin }^2}\theta } \right)d\theta }   \cr 
  & \,\,\, = 6\int_{\theta  = 0}^{2\pi } {\left( { - 36{{\sin }^2}\theta \cos \theta  - 12{{\cos }^2}\theta  + 36{{\sin }^2}\theta  + 2\cos \theta } \right)d\theta }   \cr 
  & \,\,\, = 6\left[ { - 36\int_{\theta  = 0}^{2\pi } {{{\sin }^2}\theta \cos \theta d\theta }  - 12\int_{\theta  = 0}^{2\pi } {{{\cos }^2}\theta d\theta }  + 36\int_{\theta  = 0}^{2\pi } {{{\sin }^2}\theta d\theta  + 2\int_{\theta  = 0}^{2\pi } {\cos \theta d\theta } } } \right]  \cr 
  & \,\,\, = 6\left[ { - 36\left( 0 \right) - 12\left( \pi  \right) + 36\left( \pi  \right) + 2\left( 0 \right)} \right]  \cr 
  & \,\,\, = 144\pi  \cr} $$
Next, compute the area element vector $d\bf{S}$ and $\nabla  \times {\bf{F}}$
$$\eqalign{
  & {\bf{x}} = x{\bf{i}} + y{\bf{j}} + \left( {1 - x} \right){\bf{k}}  \cr 
  & d{\bf{S}} = \left( {{{\partial {\bf{x}}} \over {\partial x}} \times {{\partial {\bf{x}}} \over {\partial y}}} \right)dxdy = \left| {\matrix{
   {\bf{i}} & {\bf{j}} & {\bf{k}}  \cr 
   1 & 0 & { - 1}  \cr 
   0 & 1 & 0  \cr 
 } } \right|dxdy = \left( {{\bf{i}} + {\bf{k}}} \right)dxdy  \cr 
  & dS = \left\| {d{\bf{S}}} \right\| = \sqrt 2 dxdy  \cr 
  & {\bf{n}} = {1 \over {\sqrt 2 }}\left( {{\bf{i}} + {\bf{k}}} \right)  \cr 
  & \nabla  \times {\bf{F}} = \left| {\matrix{
   {\bf{i}} & {\bf{j}} & {\bf{k}}  \cr 
   {{\partial _x}} & {{\partial _y}} & {{\partial _z}}  \cr 
   {xy} & {2z} & {6y}  \cr 
 } } \right| = 4{\bf{i}} - x{\bf{k}} \cr} $$
I think you had a mistake in this part $d{\bf{S}}=dS {\bf{n}}$ where $\sqrt2$ cancels. Finally, the surface integral will be
$$\eqalign{
  & I = \int\!\!\!\int {\nabla  \times {\bf{F}} \cdot d{\bf{S}}}  = \int_{x =  - 6}^6 {\int_{y =  - \sqrt {36 - {x^2}} }^{\sqrt {36 - {x^2}} } {\left( {4 - x} \right)dydx} }   \cr 
  & \,\,\,\, = \int_{x =  - 6}^6 {2\left( {4 - x} \right)\sqrt {36 - {x^2}} dx}   \cr 
  & \,\,\,\, = \int_{x =  - 6}^6 {8\sqrt {36 - {x^2}} dx}  = 8\int_{x =  - 6}^6 {\sqrt {36 - {x^2}} dx}   \cr 
  & \,\,\,\, = 8\left( {18\pi } \right) = 144\pi  \cr} $$
A: In $\iint_S {\rm curl} \> \vec{F} \cdot \vec{n} \> dS$ we have $\vec{n} = \frac{g_{x} \times g_{y}}{\left\lVert g_{x} \times g_{y} \right\rVert}$ and $dS = \left\lVert g_{x} \times g_{y} \right\rVert \,dx\,dy$ where $g$ parametrizes the surface. So, $\vec{n}\,dS = (g_{x} \times g_{y})\,dx\,dy $
That is where your mistake might be, considering the norm of the normal vector cancels out in the calculation.
