I am currently reading the paper $\textit{Classification of the actions of the circle on 3-manifolds}$, by Frank Raymond. In the proof of Lemma 2, the author enunciates the following fact without a proof: If $M^*$ is a non-compact surface with boundary, and $U^*$ is a small enough closed collared neighborhood of $\partial M^*$, then $$\text{H}^2(M^*, U^*;\mathbb{Z}) = 0.$$
The author seems to use the fact that $M^*\setminus U^*$ is non-compact and connected.
I have unsuccessfully tried to use the usual tools from a basic Algebraic Topology course for a proof of the claim. Namely, the Long Exact Sequence in Cohomology for the given pair is $$0\rightarrow \text{H}^0(M^*, U^*)\rightarrow \text{H}^0(M^*)\rightarrow \text{H}^0(U^*)\rightarrow \text{H}^1(M^*, U^*)\rightarrow \text{H}^1(M^*)\xrightarrow{\phi} \text{H}^1(U^*)\xrightarrow{\delta}\text{H}^2(M^*, U^*)\rightarrow \text{H}^2(M^*)\rightarrow \text{H}^2(U^*)\rightarrow 0.$$ The two rightmost terms vanish, since $M^*$ is a non-compact surface and $U^*$ deformation retracts onto a 1-manifold. Thus, since $\delta$ is onto, $\text{H}^2(M^*, U^*) = 0 \iff \delta = 0 \iff \phi $ is onto. Hence, it would suffice to prove that the inclusion $U^*\hookrightarrow{} M^*$ induces a surjection in 1-dimensional cohomology. This seems to have a geometric interpretation in terms of cocycles, or even 1-forms if we consider the de Rham cohomology.
The Excision Theorem is not very promising in my opinion, since there does not seem to be an appropriate way to excise a subspace from the pair $(M^*, U^*)$ so that it fundamentally simplifies. A colleague suggested to prove that the pair above has the homotopy type of a 1-dimensional CW pair, but this seems even harder to prove. By itself, this would be a very interesting fact to learn.
Another reasonable approach could be to use the fact that $(M^*, U^*)$ is a $\textit{good pair}$ in the sense of Hatcher's book, so that the quotient map $$q:(M^*, U^*)\rightarrow (M^*/U^*, U^*/U^*)$$ induces isomorphisms in homology. However, the non-compact CW space $M^*/U^*\cong M^*/\partial M^*\cong M^*\cup_{\partial M^*} C(\partial M^*)$ can be a little wild if $\partial M^*$ has infinitely many components, where $CX$ denotes the cone over $X$.
I am very curious about the two main approaches I propose, but even the smallest hint towards a proof of the original question is greatly appreciated.
You are circling very close to a correct proof. As you say, $H^2(M,U) \cong H^2(M/U)$. The space $M/U$ is homeomorphic to $M \cup \bigsqcup D^2 / \sim$, the space where we have glued on disks to the circle boundary components and $\sim$ identifies the centers of the disk to one point and collapses the boundary components homeomorphic to $\mathbb{R}$ to the same point. Since $\mathbb{R}$ is contractible, collapsing it to the distinguished point has the same effect as identifying any single point inside it with the distinguished point.
When one identifies a discrete set of points on a manifold of dimension greater than $1$, the effect on homotopy is to wedge with a number of copies of $S^1$ (that number being the cardinality minus one). Since we are picking one point from each boundary component, the set of points we identify is discrete. Wedging with copies of $S^1$ affects only the first (co)homology, so $H^2(M/U) \cong H^2(M)$. Now since $M$ is an open manifold, its top cohomology vanishes, and we are done.