Is it possible to create a completely random integer between 1 and 13 using standard dice in a D&D dice kit? I am asking if it is possible to choose a random integer using a certain set of dice. Assume for these calculations that the dice are perfectly randomly distributed. There are 7 dice in a set, with the following possible outcomes for each dice:

2, 4, 6, 8, 10, 12, and 20

You are free to interpret the outcomes as any value you like.
The dice can be rolled in any order, combination, or quantity desired.
The objective is to have 13 possible outcomes, perfectly random and uniformly distributed.
 A: This is impossible if you require that there is a fixed upper bound on the number of possible rolls: in this case, the only probabilities you can get are of the form $\frac{n}{m}$ where $m$ is a product of the numbers $2, 4, 6, 8, 10, 12, 20$, and in particular cannot be divisible by $13$. 
This is straightforward if you don't require such a fixed upper bound: as mentioned in the comments, you can discard and reroll. This strategy, like any successful strategy, necessarily has the property that it is possible that you'll have to reroll arbitrarily many times. But the probability is exponentially decaying so it's not really an issue. 
Edit: An interesting follow-up question might be to find a strategy that minimizes the expected number of rolls. Suppose you have a strategy involving $n$ rolls per iteration, with probability $p$ of succeeding each iteration (and probability $1 - p$ of needing to reroll). Then the expected number of rolls turns out to be $E = \frac{n}{p}$. So, comparing the strategies that have been suggested so far:


*

*Travis's $4$ and $10$ strategy has $n = 2, p = \frac{39}{40}$, so $E = \frac{80}{39} \approx 2.05$.

*Amitai's $20$ strategy has $n = 1, p = \frac{13}{20}$, so $E = \frac{20}{13} \approx 1.54$.

*M. Wind's "all the dice" strategy has $n = 7, p = \frac{921596}{921600}$, so $E = \frac{6451200}{921596} \approx 7.00$. 


Amitai's strategy is in fact the only possible strategy with an expected number of rolls of less than $2$ (so I take back what I said about it requiring more rolls than Travis's!), although one might go on to calculate and compare the variance of the number of rolls as well... 
A: By perfectly random, do you mean uniformly random?  Because any combination that produces 13 values (say 1d10 + (1d4-1)) would produce a (completely) random result.  Just maybe not a uniformly random result (i.e. the numbers may not show up with the same frequency).
The frequency table of the out comes of the 1d10 + (1d4-1) scheme:
$$
\begin{array}{cccccccccccccc}
\text{outcome} & 1 & 2 & 3& 4 & 5& 6& 7& 8& 9&10&11&12&13\\
\hline
\text{frequency} & .025 & .05 & .075 & .1 &.1 &.1 &.1 &.1 &.1 &.1&.075&.05&.025
\end{array}
$$
You might be able to put a penalty on the outcomes 2 to 12 and somehow produce a more uniform distribution, but likely the best  approaches would be the rejection methods specified in the comments.
A: You could use all seven dice in one throw. This gives you $2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 20 = 921600$ possible outcomes. Next you need a procedure to assign to each outcome a unique number. This allows you to distribute outcomes into $13$ groups of equal size. It is not difficult to do so.
Note that the multiple of $13$ that is closest (from below) to $921600$ is $921596 = 70892 \cdot 13$ This means that only in $4$ cases out of $921600$ you would have to reject the result and throw the dice again. The probability that this happens is $1$ in $230400$.  
A: If the goal is to minimize the expected number of dice rolled (as suggested by Qiaochu Yuan), one can improve on Amitai's method by saving the information from discarded die rolls.  Here's how it works:


*

*Roll a d20.  If the result is between 1 and 13, stop.  Otherwise roll again.

*After the second roll, we have 140 equal possibilities, with 7 choices left from the first roll, and 20 from the second roll.  If the result is one of the first 130, stop and generate a number.  Otherwise roll again.

*After the third roll, we have 200 equal possibilities, with 10 left from the second roll and 20 from the new roll.  If the result is one of the first 195 (which is a multiple of 13), stop.  Otherwise continue.
The expected number of rolls for this strategy is
$$
1 \,+\, \frac{7}{20} \,+\, \frac{10}{20^2} \,+\, \frac{5}{20^3} \,+\, \frac{9}{20^4} \,+\, \cdots \;=\; \frac{1238418740163877}{900219780219780} \;\approx\; 1.375685,
$$
where the numerators of the fractions on the left are the powers of 20 modulo 13.  The numerators repeat every 12 terms, and grouping sets of 12 terms together yields a geometric series, which is how I calculated the sum.
More generally, instead of just using a sequence of d20's, one can use any sequence of allowed dice, with $k_1$ sides for the first die, $k_2$ sides for the second die, and so forth. In this case, the expected number of rolls is
$$
1 \,+ \sum_{n=1}^\infty \frac{k_1\cdots k_n \text{ mod }13}{k_1\cdots k_n}
$$
Though I'm not entirely sure, I believe this sum is minimized for the following sequence of dice:
$$
20,\;\;8,\;\;20,\;\;20,\;\;20,\;\;8,\;\;20,\;\;20,\;\;20,\;\;8,\;\;\ldots
$$
In this case, the expected number of rolls is
$$
1 \,+\, \frac{7}{20} \,+\, \frac{4}{20\cdot 8} \,+\, \frac{2}{20^2 \cdot 8} \,+\, \frac{1}{20^3\cdot 8} \,+\, \frac{7}{20^4 \cdot 8} \,+\, \frac{4}{20^4 \cdot 8^2} \,+\, \cdots
$$
which sums to
$$
\frac{88040}{63999} \;\approx\; 1.375646\text{ rolls}.
$$
I think this may be the best possible strategy for minimizing the number of dice rolled.
A: Try rolling a D10 and a D6 with the D6 values being:
1-2 = 1
3-4 = 2
5-6 = 3
