Civitavecchia (Rome), Italy to Miami, Florida Cruise
Oceania Allura by Oceania Cruises
World25-night cruise departing November 7, 2026 from Civitavecchia (Rome), Italy
Arriving December 2, 2026 in Miami, Florida
Itinerary
- Day 1: Civitavecchia (Rome), Italy
- Day 2: Naples, Italy
- Day 3: Messina, Sicily, Italy
- Day 4: Valletta, Malta
- Day 5: Trapani, Italy
- Day 6: La Goulette, Tunisia
- Day 7: Cagliari, Sardinia, Italy
- Day 8: Cartagena, Spain
- Day 9: Valencia, Spain
- Day 10: Barcelona, Spain
- Day 11: Malaga, Spain
- Day 12: Agadir, Morocco
- Day 13: Arrecife, Lanzarote, Canary Islands
- Day 14: San Juan, Puerto Rico
- Day 15: Miami, Florida
Frequently Asked Questions
- How long is the Civitavecchia (Rome), Italy to Miami, Florida cruise?
- 25 nights on Oceania Allura.
- What cruise line operates this route?
- Oceania Cruises operates this cruise on the Oceania Allura.
- What ports does this cruise visit?
- Civitavecchia (Rome), Italy, Naples, Italy, Messina, Sicily, Italy, Valletta, Malta, Trapani, Italy, La Goulette, Tunisia, Cagliari, Sardinia, Italy, Cartagena, Spain, Valencia, Spain, Barcelona, Spain, Malaga, Spain, Agadir, Morocco, Arrecife, Lanzarote, Canary Islands, San Juan, Puerto Rico, Miami, Florida
- When does this cruise depart?
- November 7, 2026
- Where does this cruise depart from?
- Civitavecchia (Rome), Italy