Is it invalid to use a single sample to estimate more than one proportion? Lets say that I have a jar of white, black and red marbles (1000 total) but I don't know the quantity of each colour.  
I now take a sample of 10 marbles from the jar which contains 3 x white, 4 x black and 3 x red marbles.  
Is it fair for me to estimate the proportion of each colour and use the 'standard error of a proportion' for each colour independently?  For example, can I say that there are 30% white marbles with a margin error of $\sqrt{{p(1-p)}/{n\,}}$, and then do the same for black (40%) and red (30%)?
I am comfortable with the margin of error formula if we used 3 independent samples of 10 - with one experiment for each colour.  However, it feels like we are 'double-dipping' and losing statistical freedom in the estimates for black and red if we use a single sample, given that if we already know that 3 marbles are not black or red.
Apologies if I am using the wrong terminology, I am not a statistician!
 A: 
Is it fair for me to estimate the proportion of each colour and use the 'standard error of a proportion' for each colour independently? 

Yes, absolutely. It's not just fair, but also correct. 

For example, can I say that there are 30% white marbles with a margin error of $\sqrt{p(1-p)/n}$, and then do the same for black (40%) and red (30%)?

Think of it as binomial "red" vs "not red" say. The standard error of the proportion of red will be right.
However, if you start doing calculations that involve multiple colours, you must take their dependence into account. The counts for any two colours are negatively correlated.
It's not "double dipping" at all. You're just dealing with a multinomial, and if you focus on just one category vs the rest, that's binomial. All perfectly legitimate.
However, if you're talking about something akin to marbles, take care with the issue of sampling with replacement vs sampling without replacement. 
Note that if you're not sampling with replacement your standard errors don't apply because you'd have a hypergeometric not a binomial (and the multi-colour situation is multivariate hypergeometric rather than multinomial). The variance is smaller in this case, sometimes substantially smaller.
Edit: just an additional note - the margin of error isn't usually taken to be one standard deviation of the sample proportion.
