JavaScript (ES6),  142 ... 122  121 bytes

Expects an array of characters, e.g. ['5','1','2']. Returns 0 or 1.

a=>a.map(n=>D=D.map((_,i)=>D['504405076067'[d*6+i>>1]^i],p+=[w=a.length,-w,1,-1][d=D.indexOf(n)]),p=0,D=[...'254316'])|!p

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How?

The array D[] holds the face values in the following order:

 index |    0   |    1   |    2   |    3   |    4   |    5
-------+--------+--------+--------+--------+--------+--------
 face  |  front | behind |  left  |  right |   top  | bottom

We start with D = ['2','5','4','3','1','6'], which is the initial position of the die described in the challenge, according to our encoding.

The direction d of the next move is the 0-indexed position of the new top face in D[]:

 index (d) |   0   |   1   |   2   |   3
-----------+-------+-------+-------+-------
 direction |  up   |  down | right |  left

When moving towards direction d, the i-th face in the updated die is the face at the following position in the previous die:

   i = | 0 | 1 | 2 | 3 | 4 | 5
-------+---+---+---+---+---+---
 d = 0 | 5 | 4 | 2 | 3 | 0 | 1
 d = 1 | 4 | 5 | 2 | 3 | 1 | 0
 d = 2 | 0 | 1 | 5 | 4 | 2 | 3
 d = 3 | 0 | 1 | 4 | 5 | 3 | 2

This table is encoded with the following expression:

'504405076067'[d * 6 + i >> 1] ^ i

Try it online!

At each iteration, we update D[] and the position p. We test whether we're back to our starting point at the end of the process.