I've been thinking of this problem but I just can't complete a proof, here's a possible route (hopefully) towards a solution: By a simple continuity argument there is an $\displaystyle r>0$ such that on $\displaystyle 0\leq Im(z) \leq r$ we have that $\displaystyle f$ does tend to zero. Now assume the following is true
Claim: If $\displaystyle f\to 0$ on the line $\displaystyle Im(z)=s$ with $\displaystyle 0<s<1$ then $\displaystyle f\to 0$ on $\displaystyle 0\leq Im(z) \leq s$
then by an argument identical to the first $\displaystyle f\to 0$ on $\displaystyle 0\leq Im(z) \leq s+r_1$ so the set on which $\displaystyle f\to 0$ has to be $\displaystyle D$. I'm having a little trouble wih the claim though (particularly estimating $\displaystyle f$ in the verticla boundary of the set $\displaystyle 0\leq Im(z) \leq s, Re(z)>R$ for some $\displaystyle R>0$; this is enough by the MMP).
Sorry I can't be of more help, if you find a proof please post it here.
There's probably a nicer way than this (i.e. a way that invokes maximum modulus at least). But for now I don't see why this doesn't work.
Take the sequence of points $\displaystyle \left\{a_n = n+i\left(\frac{1}{n}\right)\right\}_{n=2}^\infty$. This sequence is in D.
By continuity of f we can flex the limit in and out of the function:
$\displaystyle \lim_{n \to \infty} f(a_n) = f\left(\lim_{n \to \infty} a_n\right) = f\left(\lim_{n \to \infty} n + \lim_{n \to \infty} i\left(\frac{1}{n}\right)\right) = f\left(\lim_{n \to \infty} n\right)= f\left(\lim_{x \to \infty} x\right) = \lim_{x \to \infty} f(x) = A.$
Now f is holomorphic and bounded on D, so it has no singularity at infinity (in particular, no erratic essential singularity behavior). So
$\displaystyle \lim_{z \to \infty}f(z)$ exists. So all unbounded sequences in D tend to this limit. We just showed one such sequence tends to A.
(So A is what all unbounded sequences in D tend to. In particular, any unbounded sequence with fixed imaginary component between 0 and 1 tends to A.)