Here's an attempt at the nonexistence of 3-digit strongly minimal unavoidables that have valency 3.

Suppose one exists and is made up of digits 1,2,3. Index the three permutations of the unavoidable with t=1,2,3 as well (as if the permutations were evolving over time). Say the 1s fill a pattern P at time t=1; then at time t=2, P will be filled with 2s (in position A, a subset of P) and/or 3s (in position B, i.e. P\A); then at time t=3, we must have A occupied by 3s and B by 2s (because at each "tick", all the digits must move). Pictorially:

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` | 1s | 2s | 3s |`

----+-----+-----+-----+

t=1 | A+B | | |

t=2 | | A | B |

t=3 | | B | A |

The same argument goes for the evolution of 2s and 3s through the permutations, so in general the solution must look like this, where A,B,C,D,E,F are pairwise disjoint:

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` | 1s | 2s | 3s |`

----+-----+-----+-----+

t=1 | A+B | C+D | E+F |

t=2 | D+E | A+F | B+C |

t=3 | C+F | B+E | A+D |

Now let πA be the footprint of A, i.e. the list of rows, columns and boxes it is incident with. The footprint of 1s must not change over time, so in particular we have πA+πB = πD+πE [1] where + is disjoint union of sets. But πA and πD are disjoint (look at the 3s at t=3), so [1] implies we must have πA=πE and πB=πD. But if (for example) πA=πE then we several another valid permutations of the big unavoidable set ...

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`t=4 | E+B | C+D | A+F |`

t=5 | D+A | E+F | B+C |

t=6 | C+F | B+A | E+D |

... which unfortunately clash with those at t=1,2,3 (e.g. the "A" 1s at t=1 are still there at t=5), breaking the strong minimality requirement that no permutations have any (cell,value) pair in common. So there cannot exist any 3-digit valency 3 strongly minimal unavoidables after all.