In the first place you need a parametric representation of $S$. Note that nobody would guess from the equation $x^2+y^2=1$ that this curve has length $2\pi$. So your equation n\cdot \gamma'(t)=0 is of no help.
Given the "soul" $t\mapsto \gamma(t)$ of $S$ and assuming that the curvature $\kappa$ of $\gamma$ is nonzero there is a well-defined orthonormal Frenet frame $\bigl(e_1(t),e_2(t),e_3(t)\bigr)$ along $\gamma$ with e_1=\gamma' (assuming $t$ is arc length), $e_2$ in the osculating plane, and $e_3:=e_1\times e_2$.
In terms of this frame a parametric representation of $S$ is given by
$(t,\phi)\ \mapsto\ {\bf r}(t,\phi):= \gamma(t)+r(t)\cos\phi\ e_2(t) +r(t)\sin\phi\ e_3(t)\qquad(a\leq t\leq b,\ 0\leq\phi\leq 2\pi) .$
The area $\omega(S)$ of this surface is then obtained as follows:
$\omega(S)\ =\ \int_a^b\int_0^{2\pi}\ |{\rm r}_t\times {\rm r}_\phi|\ d\phi\ dt\ .$
Because of the so-called Frenet equations for the e_i' it turns out that we can compute the integrand $|{\rm r}_t\times {\rm r}_\phi|$ without actually computing the $e_i$. The result is
|{\rm r}_t\times {\rm r}_\phi|^2=r^2\bigl(r'^2 +(\kappa r\cos\phi-1)^2\bigr)\ .
Unless r'(t)\equiv0 (a tube of constant radius) or $\kappa(t)\equiv0$ (the "soul" is a straight line) the "inner" integration $\int_0^{2\pi}\ldots\ d\phi$ will result in a complete elliptic integral, which cannot be done in elementary terms.