We try a(n almost) purely mathematical solution, without using the clues in the first part of Gareth's answer. We begin by noting that only nine different digits are specified in the long division, so we will have to infer the remaining digit from context. (If the message is the one indicated by Gareth, the D is not shown.)
From the second layer of subtractions in the tens and hundreds column of the dividend, we have
$U\equiv 2S\bmod 10$ (modulo 10 is understood from here on)
$G\equiv 2S+1$
G and U must be different; the +1 for the G comes from the fact that we had to borrow from the hundreds column to the tens column in this subtraction.
The third level subtraction in the tens column then gives either of the following, depending on whether we had to borrow from the tens column into the ones column (currently unknown):
$S\equiv 2G$
$S\equiv 2G+1$
If $S\equiv 2G+1$ and $G\equiv 2S+1$, then $S\equiv 4S+3$ from which $S=9$ and then $G=9$. This is contradictory because only one letter is allowed for each digit. Trying $S\equiv 2G\not\equiv 2G+1$ leads to $S\equiv 4S+2$, therefore $S=6$ and we infer $G=3$. We also have $U\equiv 2S=2$. Note that $G<S$, which figures into a later stage of the solution.
When these results are inserted, we may conclude from the second level subtraction that $R=9$ and from the third level subtraction that $L=4$.
So we know: $U=2, G=3, L=4, S=6, R=9$.
We now know seek letters $A$ and $O$ to complete the third level subtraction. These must add up to $6$ since $G<S$ implies no borrowing into the tens column. From the known values of other letters we may infer that $\{A,O\}=\{1,5\}$, but in which order?
The subtrahend in the third level subtraction is known to be $54132$ or $14532$; either way it must be a single digit times a four-digit divisor. The single digit, of course, is greater than or equal to $2$ for $14532$ or greater than or equal to $6$ for $54132$.
Standard divisibility tests reveal that $54132$ is divisible only by $6$ among sufficiently large digits (there are smaller factors but these do not match with a four-digit divisor). Dividing $54132$ by $6$ yields $9022$, which does not admit a four-digit multiple ending in $USS=266$. So that possibility fails and we identify the third subtrahend as $14532$; thus $O=1,A=5$.
The number $14532$ is divisible by $2,3,4,6,7$ by standard testing. Division by $7$ gives $2076$, which again fails to give a four-digit multiple ending in $266$; ditto for $14532/3=4844$. But $14532/2=7\color{blue}{266}$, and division by $4$ or $6$ will also give $7266$ as a multiple of the quotient. Thereby $F=7$ and it becomes evident that $I=8$.
We have now rendered all the digits $1$ through $9$ appearing in the subtractions and thus have most of the hidden message:
_OUGLASFIR
The first letter, corresponding to $0$, is not given in the subtractions, but the only coherent choice (at least in English) is to use the letter D making the final answer
$\color{blue}{\text{Douglas fir}}.$
This solution was obtained without completely solving the division; the blanks in the dividend, divisor and quotient are not specified. Nonetheless, the division turns out to be unique. First note that with $7266$ as one of the subtrahends, the divisor must be a factor of that number and must also give the first subtrahend, $58128$, upon multiplication by a single digit. Only $7266$ (FUSS) itself satisfies these requirements. Then from the subtrahends the digits in the quotient must be $812$ (IOU). The remainder is already known to be $2134$. From these the dividend may be computed as $5902126$ (ARDUOUS).
