I have the following idea:
Assume $X_1, X_2, \ldots$ are an i.i.d sequence of random variables with $P(X_1 = 0) = P(X_1 = 1) = \frac{1}{2}$. Then the random variable
$X = \sum \limits_{i = 1}^\infty X_i 2^{-i}$ is distributed uniformly on the interval $[0, 1]$. This follows from $P(X \le \frac{k}{2^n}) = \frac{k}{2^n}$ for any positive integers $n$ and $k \le 2^n$.
Now for a random variable $X \sim U([0, 1])$ we know that $Y := (b - a)X + a \sim U([a, b])$ and we know $\lfloor Y \rfloor \sim U(\{a, \ldots, b - 1\})$. This means we have
$$Z = \left\lfloor 10^m \sum \limits_{i = 1}^\infty X_i 2^{-i}\right\rfloor \sim U(\{0, \ldots, 10^m - 1\}).$$
Now it is obviously not very efficient to throw infinitely many coins. But for a "typical" sequence of coin flips we will know the value of $Z$ already after having thrown only finitely many coins.
For example, consider $m = 1$ and assume our first four random variables are zero. We then know
$$10 \sum \limits_{i = 1}^\infty X_i 2^{-i} = 10 \sum \limits_{i = 5}^\infty X_i2^{-i} \le 10 \sum \limits_{i = 5}^\infty 2^{-i} = \frac{10}{16} < 1,$$
from which we can deduce that $Z = 0$.
Now the difficult part is to determine after how many throws (on average) we know which value $Z$ assumes. I couldn't quite determine this, but maybe someone else has a good idea. My guess is that this method is (on average) slightly more efficient than the other methods that have been presented as answers.
Edit: Let $K$ be the minimal number of throws we need with this method. Then we know that $K \ge k$ iff there is an integer $1 \le r < 10^m$ that satisfies
$$10^m \left(\sum \limits_{i = 1}^k X_i 2^{-i} + 2^{-k}\right) > r > 10^m \left(\sum \limits_{i = 1}^k X_i 2^{-i}\right)$$
Now this is equivalent to
$$\begin{align*}
&\sum \limits_{i = 1}^k X_i 2^{k - i} + 1 > \frac{2^k r}{10^m} > \sum \limits_{i = 1}^k X_i 2^{k - i} \\
\iff{}& \sum \limits_{i = 0}^{k - 1} X_{k - i} 2^i + 1 > \frac{2^k r}{10^m} > \sum \limits_{i = 0}^{k - 1} X_{k - i} 2^i
\end{align*}$$
Now this is the case iff $\frac{2^k r}{10^m}$ is not an integer and also
$$\left\lfloor\frac{2^k r}{10^m}\right\rfloor = \sum \limits_{i = 0}^{k - 1} X_{k - i} 2^i \qquad (\star)$$
holds.
Now to get an upper estimate on $P(K \ge k)$ we can ignore the first condition (this would only make things more difficult for a miniscule improvement of our bound). Let us take a look at the event $(\star)$. The sum on the right-hand side is simply a binary representation of a (uniform) random number from $0$ to $2^k - 1$. This means the probability of $(\star)$ is equal to
$$P_k := 2^{-k} \sum \limits_{i = 0}^{2^k - 1} I\left\{i = \left\lfloor \frac{2^kr}{10^m}\right\rfloor \text{ for some } 0 < r < 10^m\right\}.$$
So how many numbers from $0$ to $2^k - 1$ can be written in this form? If $\frac{2^k}{10^m} \le 1$, we can write all numbers. However, for $\frac{2^k}{10^m} > 1$ each $r$ yields a different number. This means we get $P_k = 1$ for $k \le m \log_2(10)$ and $P_k = \frac{10^m - 1}{2^k}$ for $k > m \log_2(10)$.
Writing $c = \lfloor m \log_2(10)\rfloor$ we get the following:
$$\begin{align*}
E[K] &= \sum \limits_{k = 1}^\infty P(K \ge k) \le \sum \limits_{k = 1}^\infty P_k \le \sum \limits_{k = 1}^{c}1 + \sum \limits_{k = c + 1}^\infty \frac{10^m - 1}{2^k} \\
&= c + (10^m - 1) 2^{-c} < \lfloor m \log_2(10)\rfloor + 10^m 2^{-(m \log_2(10) - 1)} \le \lfloor m \log_2(10) \rfloor + 2
\end{align*}$$
Summarizing this method:
Let $X_n$ be the result of the $n$-th coin flip (either a $0$ or a $1$). After each coin flip, calculate the number $Z_n = 10^m\sum \limits_{i = 1}^n X_i 2^{-i}$. If $\lfloor Z_n\rfloor = \lceil Z_n + 10^m2^{-n}\rceil$ holds, then you are done and your number is given by $\lfloor Z_n\rfloor$. Otherwise, continue adding numbers until the condition holds.
The average number of coin flips of this procedure is less than $\lfloor m \log_2(10)\rfloor + 2$.
To put this into perspective: If we flip $r$ coins, we can generate a maximum of $2^r$ different values. This means that for a uniform distribution on $\{0, \ldots, 10^m - 1\}$ we need $2^r \ge 10^m$, i.e. $r \ge m \log_2(10)$. Since $\log_2(10)$ is irrational and $r$ needs to be an integer, we will always need at least $\lceil m \log_2(10) \rceil = \lfloor m \log_2(10) \rfloor + 1$ coin flips. This method will on average use only one more coin flip than this (theoretical) limit.
