I'm trying to invert the following Laplace Transform using the Cauchy Residue Theorem
$$ F(s)=\frac{p(s)}{sq(s)} $$
which is subject to the following conditions:
$$p(0)=q(0)=0$$
$$\lim_{s\rightarrow 0} \frac{p(s)}{q(s)}\hspace{0.5 cm}\text{is finite}$$
$$\lim_{s\rightarrow 0} \frac{p(s)}{s}\hspace{0.5 cm}\text{is finite}$$
$$\lim_{s\rightarrow 0} \frac{p(s)}{sq(s)}=\infty$$
The function $q(s)$ has several other zeroes along the negative real axis, which I can easily handle.
The questions I have are:
Is $s=0$ is a pole, and if so, am I correct in concluding that it's a pole or Order 1?
If it is indeed a first order pole, how do I evaluate the residue?"
Assuming your function $p,q$ are holomorphic, then yes, $s=0$ is a pole of $F(s) = \frac{p(s)}{sq(s)}$, since $\lim\limits_{s\rightarrow 0} F(s) = \infty$. It would be an essential singulartiy, if there additionally existed a sequence $(s_n)_{n\in\mathbb{N}}\subset\mathbb{C}$ with $\lim\limits_{s\rightarrow 0}s_n=0$ and $\lim\limits_{n\rightarrow\infty} F(s_n)\neq \infty$.
Further, it has order $1$, since the function $s\cdot F(s) = \frac{p(s)}{q(s)}$ is holomorphic around $s=0$.
One can calculate the residue of a $m$-th order pole $s_0\in\mathbb{C}$ by $$\operatorname{Res}(F,s=s_0) = \frac{1}{(m-1)!}\left.\frac{d^{m-1}}{ds^{m-1}}\right|_{s=s_0} F(s)\cdot(s-s_0)^m,$$
so in your case it is just $\operatorname{Res}(F,s=0) = \left.\left(\frac{p(s)}{sq(s)}\cdot s\right)\right|_{s=0} = \left.\frac{p(s)}{q(s)}\right|_{s=0}$.