I'm assuming $n$ is a power of $2$ and that $q$ is an odd prime larger than $n$. I'm discarding the trivial case $s_1 = s_2$.
If you consider everything $\mod q$, then it is most likely over the choice of $a$ that there exists $s_1 \neq s_2$ such that $\|a s_1 - a s_2\| = \sqrt{n}$. Indeed, $a$ is invertible in $R_q$ with probability about $1 - n/q$. Take $s_2 = s_1 - a^{-1}$, then you have $a s_1 - a s_2 = 1 \mod q$ and the embedding norm of $1$ is $\sqrt{n}$.
If you do not consider this $\mod q$, i.e. you work in $R=\mathbb Z[x]/⟨f(x)⟩$, then you are precisely asking for the minimal distance $\lambda_1(\mathfrak I)$ of the ideal lattice $\mathfrak I$ generated by $a$. For such an ideal lattice, we can estimate rather precisely this minimal distance. A simple lower bound is
$\lambda_1(\mathfrak I) \geq \Delta_K^{1/2n} \cdot N(a)^{1/n}$, where $N$ denotes the algebraic norm of $a$ (that is, the product of all its embeddings), and $\Delta_K$ is the discriminant of field $K = \mathbb Q(x)/(x^n+1)$. The reason is that the minimal vector $x$ must generate a subideal of $\mathfrak I$, so $N(x) \geq N(a)$, and $\|x\|^n \geq \Delta_K^{1/2} N(x)$. An upper bound is also given by Minkowski's theorem.