$f$ has no zeros on the boundary of the unit circle $\Bbb D$, therefore
$$
\varepsilon_0 = \frac{ \min \{ |f(z)| : |z| = 1 \}}{1 + \max \{ |g(z)| : |z| = 1 \} } \, .
$$
is strictly positive.
Then for $0 \le \varepsilon \le \varepsilon_0$ and $|z| = 1$,
$$
| (f(z) + \varepsilon g(z)) - f(z) | = \varepsilon |g(z)| < |f(z)|
$$
and it follows from Rouché's theorem that $f(z) + \varepsilon g(z)$ and
$f(z)$ have the same number of zeros in $\Bbb D$. Since $f$ has exactly
one (simple) zero, it follows that $f(z) + \varepsilon g(z)$ also
has exactly one zero $z_\varepsilon$ in the unit disk.
The solution $z_\varepsilon$ of $f(z) + \varepsilon g(z) = 0$
can be represented as
$$
z_\varepsilon = \frac{1}{2 \pi i} \int_{\partial \Bbb D}
\frac{z (f'(z) + \varepsilon g'(z))}{f(z) + \varepsilon g(z)} \, dz
$$
(Proof: For fixed $\varepsilon$, $f(z) + \varepsilon g(z) =
(z - z_\varepsilon)h(z)$ where $h$ is holomorphic and not zero
in $\Bbb D$. Then
$$
\frac{z (f'(z) + \varepsilon g'(z))}{f(z) + \varepsilon g(z)}
= z \frac{h'(z)}{h(z)} + 1 + \frac{z_\varepsilon}{z_\varepsilon - z}
$$
and that has exactly one (simple) pole in $\Bbb D$, with residue $z_\varepsilon$.)
Since the integrand is continuous as a function of
$(z, \varepsilon) \in \partial \Bbb D \times [0, \varepsilon_0]$
it follows that the integral is a continuous function of $\varepsilon $.
(It is even an analytic function of $\varepsilon $ if we consider
complex $\varepsilon $ with $|\varepsilon| < \varepsilon_0 $.)
An alternative proof would be to apply the implicit function theorem
to $F(\varepsilon, z) = f(z) + \varepsilon g(z)$, viewed as a function
from $\Bbb R \times \Bbb R^2 \to \Bbb R^2$.
Writing $f(z) = u(x, y) + i v(x,y)$, the derivative of $F$
with respect to $(x, y)$ at $(0, (0, 0))$ is
$$
\begin{pmatrix}
u_x(0, 0) & u_y(0, 0) \\
v_x(0, 0) & v_y(0, 0)
\end{pmatrix} =
\begin{pmatrix}
u_x(0, 0) & u_y(0, 0) \\
-u_y(0, 0) & u_x(0, 0)
\end{pmatrix}
$$
which is invertible because its determinant
$$
u_x(0, 0)^2 + u_y(0, 0)^2 = |f'(0)|^2
$$
is not zero.
Since this seems to be a popular question, here is my full-credit solution:
Consider the following function
$$g(z) = \begin{cases}
f(z) & \text{ if }|z|\leq 1 \\
\frac{1}{\overline{f(1/\overline{z})}} & \text{ if }|z|\geq 1
\end{cases}
$$
Note that $g$ is well defined since $\frac{1}{\overline{z}} = z$ for $z$ such that $|z| = 1$, $|f(z)| = 1$ whenever $|z| = 1$, and
$$\frac{1}{\overline{f(1/\overline{z})}} = \frac{1}{\overline{f(z)}} = f(z)$$
We now verify that $\frac{1}{\overline{f(\frac{1}{\overline{z}}})}$ is holomorphic for $|z|>1$. Since $f(z)$ is holomorphic and non-vanishing on $D$, we know that $\frac{1}{f(\frac{1}{z})}$ is holomorphic for $z \in \mathbb{C}$ such that $|z|>1$ as it is the composition of holomorphic functions. Thus, by Theorem 4.4 in Chapter 2 of Stein, for every $z_0 \in \mathbb{C}$ such that $|z_0|>1$, there is a disc centered at $\overline{z_0}$ such that $\frac{1}{f(\frac{1}{z})} = \sum_{n=0}^\infty a_n(z-\overline{z_0})^n$. Then,
$$\frac{1}{\overline{f(\frac{1}{\overline{z}}})} =\sum_{n=0}^\infty\overline{a_n}(z-z_0)^n$$
Thus, $\frac{1}{\overline{f(\frac{1}{\overline{z}}})}$ is holomorphic on $\big\{z \in \mathbb{C}:|z|>1\big\}$. We claim that $\int_T g(z)dz =0$ for every triangle in the complex plane. By Cauchy's Theorem, we know that if a triangle is completely inside $D$ or $\big\{z:|z|>1\big\}$, $\int_Tg(z)dz = 0$. Consider now a triangle as that depicted below:
Zooming in on the region of intersection,
We then see by Cauchy's theorem that the only contributor to $\int_T g(z)dz$ is the following contour
Note here that the treatment of the case where the triangle touches the unit disk at one point is in the proof Stein provides of the Symmetry Principle in Chapter 2. If we continue this process for both triangles in the above diagram, and continue doing so for those that follow, the resultant limiting line integrals will be zero. Thus, we conclude that
$$\int_T g(z)dz = 0$$
for all triangles in the complex plane. Thus, by Morera's Theorem, $g$ is holomorphic in $\mathbb{C}$. Furthermore, as $g$ is clearly bounded, we know by Liouville's Theorem that $g$ is constant. We conclude then that $f$ is constant.
For those who potentially need it, there are lots of free copies of Stein's text online. Here is a link that I use: Complex Analysis
Best Answer
When doing this problem, I really wanted the condition that $\hat{f}(\xi) = 0\; \forall\; x \in \mathbb{R}$. Indeed, by pulling out the textbook, this is the condition given for the problem.
First, we get agreement of $A$ and $B$ on the real axis since $\hat{f}(\xi) = 0\; \forall \xi \in \mathbb{R}$. We then consider the extension $F$, which is bounded (since $f$ is of moderate decrease) and entire except possibly on the real-axis. Also note that $F$ is continuous everywhere since $A$ and $B$ agree on the real-axis. Then we have that the form $f(z)\; dz$ is closed and, by Morera's Theorem, $F$ is entire. So, $F$ is both bounded and entire, whence, by Liouville's Theorem, $F$ is constant. But $|F(yi)| = |A(yi)| < \varepsilon$ for $y$ sufficiently large and so $F \equiv 0$. It then follows that $A(0) = 0$ and so by FTC, $f \equiv 0$.
I remark that I initially thought to borrow your idea regarding Theorem 3.5, but we don't know that $\hat{f}$ is of moderate decrease. So this method seems to be a dead end.