Very nice problem! We will show that taking all $a_{i}$ between $-\frac{p}{2}$ and $\frac{p}{2}$ congruent to $p \pmod 3$ makes the product a power of 3. For example, for $p = 11$, we take $\{-4,-1,2,5\}$ as the $a_{i}$ so the product becomes \[\frac{15}{4}\cdot \frac{12}{1}\cdot \frac{9}{2}\cdot \frac{6}{5}= 3^{5}.\] Note that the denominator, the product of the $|a_{i}|$, is the product of all positive integers less than $\frac{p}{2}$ that are not multiples of 3. The numerator is the product of all the multiples of 3 from $\frac{p}{2}$ to $\frac{3p}{2}$. In other words, it is a power of 3 times all the positive integers from $\frac{p}{6}$ to $\frac{p}{2}$. However, this is exactly the numerator -- for every multiple of 3 between $\frac{p}{6}$ and $\frac{p}{2}$, we can divide out all powers of 3 to get a positive integer less than $\frac{p}{6}$, and conversely, every positive integer $\frac{p}{6}$ can be matched with a multiple of 3 between $\frac{p}{6}$ and $\frac{p}{2}$. Thus, we are left with a power of 3, as desired...

A different solution: We will construct $a_i$ recursively. Define $b_0 = 1$. Now having defined $b_i$, choose a $a_{i+1} \in \left( \frac{-p}{2} , \frac{p}{2} \right)$ such that $\frac{p-a_{i+1}}{b_0}$ is a power of $3$ (it is easy to see such a $a_{i+1}$ exists, by looking at $b_i,3\cdot b_i,3^2 \cdot b_i,\ldots$) and define $b_{i+1} = a_{i+1}$. By induction we have that $$\prod_{j=1}^{i+1} \frac{p-a_j}{|a_j|} = \frac{\text{power of }3}{b_{i+1}}$$Now by PHP, we will have $b_m = b_n$ for some $m < n$. But then $$\prod_{j=m+1}^n \frac{p-a_j}{|a_j|} = \text{power of }3 $$So we are done. $\blacksquare$