Edit: I think I can back up Oray's solution with a proof and confirm that the answer is indeed
$24$
His algorithm suffices to give the upper bound so it only remains to show that it is also a lower bound.
Lower bound
Call a pair of words 'neighbours' if they differ by one letter e.g, $ABCD$ and $ABAD$ are neighbours.
Let us consider the map $A \rightarrow 0, B \rightarrow 1, C \rightarrow 2, D \rightarrow 3$ so that each word maps to a point in 4-dimensional space, e.g, $BACD \rightarrow (1, 0, 2, 3)$. Then the set of words is a finite hypercube lattice (of side length 3) and the set of neighbours of a given word, $W$, is the set of points which lie on some axis-parallel line passing through the point corresponding to $W$.
The set of words beginning with a given letter is a finite cubic lattice in this space.
Key Observation: Each guess eliminates $10$ possible answers from the cube containing the guess and $1$ possible answer from each of the other cubes.
Now, suppose that the number of guesses we make is less than $24$. Then there is at least one cube which contains at most $5$ guesses.
We may ask the following question: what is the maximum number of words we can eliminate from a single cube with $5$ guesses?
I propose that the answer is $45$.
To show this, notice that each cube further subdivides into four planes corresponding to fixing the second letter of the word. With $5$ guesses, there is at least one of these planes which contains at least two guesses. This pair of guesses will share at least two neighbours (call them intersections of the guesses). Alternatively, we can subdivide the cube lattice into planes in the other two directions (defined by fixing the third or fourth letter) and use the same logic to show that there exists a plane, in each of these directions, which contains at least two guesses with at least a pair of intersections. On face value this seems to constitute six intersections but really there are three possible scenarios to consider.
(i) These intersections may be distinct, in which case there are a total of six intersections.
(ii) One or more of these intersections may appear twice, in which case any double intersection counts twice (as it prevents two guesses from eliminating an additional possible answer). So again there are essentially a total of six intersections.
(iii) One intersection may appear three times. This is a special case as the triple intersection may only count twice (it prevents two guesses from eliminating a possible answer but if these two guesses also intersect here it is to neither one's detriment). Hence, in this case there are essentially five intersections.
The point is that the maximum number of words in a given cube that can be eliminated as possible answers by $5$ guesses in that cube is $5 \times 10 - 5 = 45$. The maximum number of words in a given cube that can be covered by $18$ guesses in other cubes is $18$. Hence, if we are limited to at most $23$ guesses, then there is some cube with at most $45 + 18 = 63$ out of $64$ words eliminated as possible answers. Hence, the number of guesses we must make is at least $24$.
Note 1: If the above logic does not convince you, it does not take very much computing power to show that, in a given cube of words with $5$ guesses, we can eliminate at most $45$ possible answers since ${{64}\choose{5}} = 7,624,512$ which is not terribly big.
Note 2: It should be fairly clear that it's not worth having less than five guesses in a given cube if we are limited to $23$ overall since we make up only one guess by guessing in a different cube and lose at least that by not guessing in the given cube.
Note 3: One thing that falls out from this analysis is that it may be possible, with $23$ guesses, to guarantee that either you win the prize or you definitely know your friend's word. I don't know if this is achievable but would be very interesting to see if so.
