Yes, you get a deformation retract. More formally...
Suppose $M$ is a topological manifold, $N\subseteq \partial M \subseteq M$ is a component of $\partial M$ and that the inclusion $N\rightarrow M$ is a homotopy equivalence. Then $M$ deformation retracts to $N$.
Here's the idea of the proof:
From Hatcher Algebraic Topology, Corollary 0.20 (pg. 16 in my copy), it's enough to show that $(M,N)$ has the homotopy extension property.
Now, boundary components in topological manifolds are known to have collar neighborhoods (see Morton Brown, "Locally flat imbeddings of topological manifolds", Annals of Mathematics, Vol. 75 (1962), p. 331-341.) This is more well-known in the smooth category, where it is also much easier to prove.
So, some closed neighborhood of $N$ in $M$ is of the form $N\times [0,1] = N\times I$, where we are identifying $N$ with $N\times \{0\}\subseteq M$..
Now suppose $f:M\rightarrow X$ is a continuous function (where $X$ is any topological space). Suppose $F:N\times I\rightarrow X$ has the property that $F(n,1) = f(n)$. We wish to find a continuous function $G:M\times I\rightarrow X$ satisfying two conditions. First, for any $n\in N$, $t\in I$, that $G(n,t) = F(n,t)$. Second, for any $m\in M$, $G(m,1) = f(m)$.
To that end, set $G(m,t) = \begin{cases} f(m) & m\notin N\times [0,1]\\ F(n, \max\{s,t\}) & m=(n,s)\in N\times [0,1].\end{cases}$
To check the first condition, note that $n\in N\cong N\times \{0\}$ means the corresonding $s$ is $s= 0$. Since $t\in[0,1]$, $\max\{s,t\} = \max\{0,t\}= t$, so $G(n,t) = F(n,t).$
To check the second condition, we have $G(m,t) = f(m)$ if $m\notin N\times [0,1]$, regardless of the value of $t$. On the other hand, if $m = (n,s)\in N\times [0,1]$, then because we are taking $t=1$, we have $\max\{s,t\} = t$. So $G(m,1) = F(m,1) = f(m)$.
Finally, we must verify that $G$ is actually a continuous function. Since $f,F,$ and $\max$ are all continuous, we see that $G$ is a continuous funciton ... as long as it is a function. That is, we still need to verify that $G$ is well defined at the "transition" points where $s = 1$.
But, if $m\in N\times [0,1]$ is of the form $m = (n,1)$, then $s=1$, so $\max\{s,t\} = s=1$, so $G(m) = F(n, \max\{s,t\}) = F(n,1) = f(n).$
Best Answer
Your understanding is pretty much right, but I would add a few things.
For $N$ to be a tubular neighborhood, it's not enough to require that $N$ is open and its closure is a manifold with boundary (you meant $\bar{N}$, not $\bar{A}$, right?). You also want it to be nicely imbedded locally into $X$. Wikipedia and Mathworld both describe this pretty well.
Any tubular neighborhood (not just one) should have the deformation retract property you mention. It seems conventional to assume that $X$ is a manifold and $A$ is a submanifold, in which case existence is known. I don't know about more general existence theorems, but there is a similar concept called a "regular neighborhood" of a subcomplex in a piecewise linear manifold. See, for example, Hempel's book on three-manifolds.