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
You want to show that $f(t, x) = (1 - t)x + tx/|x|$ is not in $A$ for any $t\in [0, 1]$ and any $x\notin A$.
Suppose it is not the case. Then there is $x\notin A$ and $t\in [0, 1]$ such that $f(t, x) = \lambda x \in A$, where $\lambda\in \Bbb R$ is the number $1 - t + t/|x|$.
If $|x| > 1$, then we have $\lambda \geq (1 - t)/|x| + t/|x| = 1/|x|$, hence $|f(t, x)| \geq 1$. This contradicts the assumption that every point in $A$ has norm $< 1$.
If $|x| \leq 1$, then we have $\lambda \geq 1 - t + t = 1$. But we also have $0\in A$. By convexity of $A$, the linear combination $1/\lambda \cdot f(t, x) + (1 - 1/\lambda) \cdot 0 = x$ is in $A$. This contradicts the assumption $x\notin A$.