Anything is possible and lowering the lightship weight (LS) of ships will save fuel, but the key issue would be any change must not reduce the seaworthiness of ships.
This question is very broad.
Regulations have been established for commercial steel ships (DNV GL, Lloyds, ABS, individual countries). These regulations take into account the harsh marine environment. In many cases, they represent the hard learned lessons from accidents like the Titanic. Regulatory agencies, would have to be overcome. It is on thing to do it for small craft, another for commercial ships.
Safety (which probably has more to do with regulations).
Fire: Steel melts at around 1370°C. Plastics melt at temperatures below 300°C. A candle burns at 1000°C. A typical fire in a steel compartment can be extinguished by closing off the air supply. On a plastic hulled vessel, the hull would burn (which would probably extinguish the fire, but cause stability problems).

Strength: A ship is built from the keel up. The hull steel plates and stiffeners must displace the displacement weight of the ship. The hull must be able to withstand hurricane force wind and waves. Any plastic would not have the strength of steel, so more structure would have to be used, so any loss in hull weight would probably be offset in structure weight. A steel ship already flexes in heavy seas. The longer the ship, the greater the flex.

Impact: There are a number of estimations for steel weight estimation. They are summarized in Preliminary parametric estimation of steel weight for new ships. They report a 6.25% Error for a 8500TEU container ship when using the following formula:
$$ Steel Weight = 0.034 \times L^{1.7}\times B^{0.7}\times D^{0.4}\times C_B^{1.7} $$
$$ Steel Weight = 0.034 \times {317m}^{1.7}\times {45.8m}^{0.7}\times {25}^{0.4}\times {0.6}^{0.5} = 24, 780mt$$
Assume half is structural and half hull or 12,390mt.
From Electrical Systems and Engine Room: Displacement, LWT and DWT
$$Displacement = C_B \times L \times B\times D\times \rho_{SW}$$
$$Displacement = 0.6 \times 317m \times 45.8m\times 25\times 1.025 tonnes/m^3 = 223,224mt$$
These are just estimates, but at best with no increase in structure and if plastic had no weight, you'd save < 5.5%.