Mathieu Baillif provided an answer to the question. I'll leave here some more notes for future readers. The following strengthening of the theorem in Gauld's book "Non-metrizable manifolds" holds:

Let $X$ be a Hausdorff space, and $X' \subset X$. Then $X'$ is collared in $X$ and $X'$ is strongly paracompact in $X'$ if and only if $X'$ is closed in $X$, $X'$ is strongly paracompact in $X$, and $X'$ is locally collared in $X$.

Suppose then that $X$ is a manifold-with-boundary; e.g Hausdorff and locally upper-Euclidean and not necessarily connected. It can be shown that $\partial X$ is closed in $X$, and $\partial X$ is locally collared in $X$. It can also be shown that in a locally compact space, which a manifold-with-boundary is, strong paracompactness is equivalent to paracompactness. Therefore, we have the following theorem:

Let $X$ be a manifold-with-boundary. Then $\partial X$ is collared in $X$ and $\partial X$ is paracompact in $\partial X$ if and only if $\partial X$ is paracompact in $X$.

This is quite a nice theorem. However, it leaves open the characterization of of non-paracompact collared boundaries. I would be interested to hear if someone can complete the characterization.

The following example shows that there exists non-paracompact collared boundaries. The open long ray is a non-paracompact manifold. Therefore, its cartesian product with the interval $[0, 1]$ is a manifold-with-boundary whose boundary is non-paracompact and clearly collared.

Some background

Here are some more properties of collared subsets, which I think to have proved:

• If $X' \subset X$ is $X$-collared, then $X' \times Y$ is $(X \times Y)$-collared. Applied to manifolds-with-boundary, if $X$ is a manifold-with-boundary, $Y$ is a (boundaryless) manifold, and $\partial X$ is $X$-collared, then $X \times Y$ is a manifold-with-boundary, and $\partial (X \times Y) = \partial X \times Y$ is $(X \times Y)$-collared.
• If $X'_i \subset X_i$ is $X_i$-collared for each $i \in I$, then $\sqcup X'_I$ is $\sqcup X_I$-collared, where $\sqcup$ denotes disjoint sum.

There are two distinct definitions of a collar, the one used in Brown (Locally flat imbeddings of topological manifolds), and the one used in Connelly (A new proof of Brown's collaring theorem). By a collar I refer to Connelly's definition. Existence of Connelly-collar implies existence of Brown-collar. The converse holds in metric spaces (e.g. paracompact manifolds), but not in general. The definition of Connelly-collar is:

A Connelly-$X$-collar of $X' \subset X$ is a function $h : X' \times [0, 1] \to X$, such that

• $h$ is a closed embedding,
• $h(x, 0) = x$, for each $x \in X'$,
• $h(X' \times [0, 1))$ is open in $X$

In this case, say that $X'$ is $X$-collared. Since $h$ is closed, we see that $X'$ is closed in $X$.

A Brown-$X$-collar of $X' \subset X$ is an open embedding $g : X' \times [0, 1) \to X$ such that $g(x, 0) = x$ for each $x \in X'$. Given a Connelly-collar $h$, we have that $h|(X' \times [0, 1))$ is a Brown-collar.