Every noncompact manifold admits nonzero vector fields, or more generally,
vector fields with any specified set of isolated zeros along with the behavior near
that zero.

However, if you have information of the behavior of a vector field
near infinity, or just in a neighborhood of the boundary of a compact set, there
is an index theorem. Perhaps this is the case with your $\mathbb T^! \times \mathbb R$.

In the particular case of a cylinder, there is a simple way to calculate the index.
Take any compact subcylinder delimited by two circles. Map the cylinder to the plane
minus the origin. Around each of the curves, the vector field has a turning number:
as you go around the curve counterclockwise, the vector field turns by some number
of rotations (counting counterclockwise as positive. The index of the
vector field in the compact subannulus is the difference: the number of turns on
the outer boundary minus the number of turns on the inner boundary.

One way to describe a general formula is this:
let $N^n$ be manifold, and let
$M, \partial M \subset N$ be a compact submanifold. Let $X$ be a vector field that is
is nonvanishing in a neighborhood of $\partial M$.

Choose an outward normal vector field $U$ along $\partial M$; now arrange $X$ so
that its direction coincides with $U$ only in isolated points, so if we project
$X$ to $N$ along $U$, it is a vector field with isolated singularities. Let $i_+(X)$
be the sum of the Poincaré-Hopf indices over all singularities where $X$ is oriented
outward. Then the Poincaré-Hopf index $i(X)$ of $X$ in $M$ equals the Euler characteristic
of $M$ minus $i_+(X)$.

Here's one proof: triangulate a neighborhood of $M$ so that $\partial M$ is a subcomplex, and so that $X$ is *transverse* to the triangulation except
near the singularities, in the sense that in any simplex, the foliation defined by
$X$ is topologically equivalent to the kernel of a linear map in
general position of the simplex to $\mathbb R^{n-1}$. Put a $+1$ at the barycenter
of each triangle of even dimension, and a $-1$ at the barycenter of each triangle
of odd dimension. Think of $X$ as a wind that blows these numbers along, so
that after an instant, all numbers (except for exceptions near the zeros of $X$)
are inside an $n$-simplex. In any typical simplex, all the signs cancel out.
However, along the boundary, some of the numbers are blown away and lost.
To regularize the situation, modify $X$ by pushing in the negative normal direction.
Now $X$ points inward everywhere except in a neighborhood of points where
it coincides with the outward normal. Thus everything cancels out except for
local contributions given by $i(X)$ and $i_+(X)$.