I tried doing this and didnt account for the width of each strip so I ended up with the formula:
I ended up with 4πR instead of 4πR^2
My thought was that since integrals are summing over all the 1 dimensional rings adding them all should give the surface area,
Could you explain to me why we need to account for this infinitesimal width and why it makes such a big difference?
You can never get an area adding lengths. You need to add infinitesimal areas of "slices" of the sphere, which are, up to higher order infinitesimals (hence not contributing to the integral), lateral surfaces of truncated cones. The "slice" corresponding to the angle $\theta$ is the lateral surface of a truncated cone with "radius" $R\sin\theta$ and generator (or infinitesimal slant height) equal to the length of an arc of the circle of radius $R$ and central angle $d\theta$, which is $Rd\theta$. Therefore, the surface area is $2\pi R^2\sin\theta d\theta$. Adding those infinitesimal contributions gives you the desired area.
(Observe that the surface area of a truncated cone of radii $r$ and $R$ and slant height $g$ is $\pi(R+r)g$, but when $R$ and $r$ are infinitesimally close it becomes $2\pi R g$).