As you say yourself, there will be a dependence on time and you'll need a differential equation to solve the problem. Let's make some simplifying assumptions. We'll assume that our body is spherical and has a radius $R$. We'll assume that its density is homogeneous and that it is denser than the fluid in which it will be immersed. We'll assume it drops from a height $h_0$ from the surface of the liquid, i.e. its center of mass is at height $h_0$. We'll neglect friction and splashing and any phenomenon that will complicate our analysis. We'll only take gravity and the Archimedean force into account.
The variable component here is the Archimedean force which depends on the immersed volume which will change in time. The formula for this force is
$$F_A=\rho_{f} g V(h)$$
where $\rho_{f}$ is the fluid density, $g$ is the gravitational acceleration in the vicinity of the Earth's surface and $V(h)$ is the immersed volume. The formula for the immersed volume is
$$V(h)=\begin{cases}0 \; \text{ if } h\geq R \\ \frac{\pi}{3}(R-h)\left[2R^2+Rh+h^2\right] \; \text{ if } -R<h<R \\ \frac{4}{3}\pi R^3 \; \text{ if } h\leq -R \end{cases}$$
We therefore have the following equation of motion
$$-g\left[\rho_0 \frac{4}{3}\pi R^3-\rho_f V(h)\right] = \rho_0 \frac{4}{3}\pi R^3 \frac{d^2 h}{dt^2}$$
Where $\rho_0>\rho_f$ is the density of our sphere. This is a second order nonlinear ODE because $V(h)$ is of third order in $h$ for part of its range.
Introducing the new variables $x=h/R$ and $\tau=\sqrt{g/R}t$ and a new function
$$P(x) = \begin{cases}0 \; \text{ if } x\geq 1 \\ \frac{2-x-x^3}{4} \; \text{ if } -1<x<1 \\ 1 \; \text{ if } x\leq -1 \end{cases}$$
we can rewrite our equation as
$$\frac{d^2x}{d\tau^2}=\frac{\rho_f}{\rho_0}P(x)-1$$
We can integrate this equation w.r.t. $x$ once to reduce the order of our equation. This gives us
$$\frac{1}{2}\left(\frac{dx}{d\tau}\right)^2=(h_0/R-x)+\frac{\rho_f}{\rho_0}Q(x)$$
in which I introduced $Q(x)$ which is the integral of $P(x)$,
$$Q(x) = \begin{cases}0 \; \text{ if } x\geq 1 \\ \frac{-5+8x-2x^2-x^4}{16} \; \text{ if } -1<x<1 \\ x \; \text{ if } x\leq -1 \end{cases}$$
We can therefore solve for the time $\tau$ in function of $x$
$$\tau=\frac{1}{\sqrt{2}}\int_{-1}^{h_0/R}\frac{dx}{\sqrt{h_0/R-x+\frac{\rho_f}{\rho_0}Q(x)}}$$
I've computed a first order approximation in $\rho_f/\rho_0$ for the case of the sphere dropping from just above the surface until the moment its gets fully immersed:
$$\tau = 2+\frac{43}{210}\frac{\rho_f}{\rho_0}+O((\rho_f/\rho_0)^2)$$
or
$$t=2\sqrt{\frac{R}{g}} +\frac{43}{210}\frac{\rho_f}{\rho_0}\sqrt{\frac{R}{g}}+O((\rho_f/\rho_0)^2)$$
The zeroth order term is what you'd expect if there was no fluid, the next term gives a correction for a very dense sphere compared to the density of the fluid.
