If you have a binomial random variable $X$, of size $N$, and with success probability $p$, i.e. $X \sim Bin(N;p)$, then the mean of X is $Np$ and its variance is $Np(1-p)$, so as you say the variance is a second degree polynomial in $p$. Note however that the variance is also dependent on $N$ ! The latter is important for estimating $p$:
If you observe 30 successes in 100 then the fraction of successes is 30/100 which is the number of successes divided by the size of the Binomial, i.e. $\frac{X}{N}$.
But if $X$ has mean $Np$, then $\frac{X}{N}$ has a mean equal to the mean of $X$ divided by $N$ because $N$ is a constant. In other words $\frac{X}{N}$ has mean $\frac{Np}{N}=p$. This implies that the fraction of successes observed is an unbiased estimator of the probabiliy $p$.
To compute the variance of the estimator $\frac{X}{N}$, we have to divide the variance of $X$ by $N^2$ (variance of a (variable divided by a constant) is the (variance of the variable) divided by the square of the constant), so the variance of the estimator is $\frac{Np(1-p)}{N^2}=\frac{p(1-p)}{N}$. The standard deviation of the estimator is the square root of the variance so it is $\sqrt{\frac{p(1-p)}{N}}$.
So , if you throw a coin 100 times and you observe 49 heads, then $\frac{49}{100}$ is an estimator of for the probability of tossing head with that coin and the standard deviation of this estimate is $\sqrt{\frac{0.49\times(1-0.49)}{100}}$.
If you toss the coin 1000 times and you observe 490 heads then you estimate the probability of tossing head again at $0.49$ and the standard devtaion at $\sqrt{\frac{0.49\times(1-0.49)}{1000}}$.
Obviously the in the second case the standard deviation is smaller and so the estimator is more precise when you increase the number of tosses.
You can conclude that, for a Binomial random variable, the variance is a quadratic polynomial in p, but it depends also on N and I think that standard deviation does contain information additional to the success probability.
In fact, the Binomial distribution has two parameters and you will always need at least two moments (in this case the mean (=first moment) and the standard deviation (square root of the second moment) ) to fully identify it.
P.S. A somewhat more general development, also for poisson-binomial, can be found in my answer to Estimate accuracy of an estimation on Poisson binomial distribution.
Best Answer
"Better" depends on context and purpose. Before addressing this issue, though, let's consider the data.
As a point of departure we might assume--hypothetically, being willing and happy to be proven wrong later in the analysis--that the outcomes of each subject's attempt at the task are independent. This permits us to hold up a simple model for scrutiny, one in which each subject $i$ has a constant chance $p_i$ of success with each attempt. It follows that the raw counts of successes $(x_i,\ i=1, 2, \ldots, n)$ consist of six (or, more generally, $n$) independent realizations of Binomial$(m, p_i)$ variables $X_i$ (with $m=20$ in this case). In this case the raw counts are $(19,16,16,13,18,14)$, obtained by multiplying the reported success rates by $20$.
This is a complicated model because it has as many parameters ($n$ of them) as there are data. To see whether the complication is worthwhile, we ought to compare this model to a simplified version. The simplest is that all the $p_i$ are equal: the subjects have equivalent abilities at the task. Is there a small set of simple, easily understood, summary statistics that might help give us some quick insight into which model would be appropriate?
In the spirit of an Analysis of Variance we might be inclined to compare the variance of the dataset--which will comprise the variances inherent in each of the $X_i$ together with the variance of the $p_i$--to some measure of the variance to be expected when all the $p_i$ are equal. Therefore we compute:
The mean of the $p_i$ is $p = (1/n)\sum_{i=1}^n p_i.$ There are several ways to estimate this, but one of the simplest--as justified by the hypothesis that all the $p_i$ are equal--is the sample mean,
$$\hat{p} = \frac{1}{n}\sum_{i=1}^n \frac{x_i}{m} = \frac{4}{5} = 0.8.$$
The variance of each $X_i$ is $m p_i(1-p_i)$; under the hypothesis of equality, this is $m p(1-p)$, which can be estimated as
$$\hat{\sigma} = m \hat{p} (1 - \hat{p}) = \frac{16}{5} = 3.2.$$
The variance of the data is
$$\text{Var}(x_i) = \frac{1}{n}\sum_{i=1}^n (x_i - m \hat{p})^2 = \frac{13}{3} = 4.\bar{3}.$$
Please notice that, in the spirit of description and exploration, division by $n-1$ in this variance calculation could be considered irrelevant. However, should one feel a need to so change the denominator, the result would be $26/5 = 5.2$.
The statistic (3) can be understood as arising from two components: the variation in subject performances due to chance plus the variation in capabilities between the subjects. That is why the two standard deviations computed in the question (which are the square roots of (2) and (3)) may differ. It becomes clear that they work together to give two separate pieces of information about the data.
It is attractive to take one more step. ANOVA teaches us that the relevant statistic to examine would be the ratio
$$\text{Var}(x_i) / \hat{\sigma}.$$
A value much greater than $1$ would indicate the $p_i$ should be treated as non-constant. In the present case--using the alternative expression for the variance employed in the question--this ratio equals $(26/5)/(16/5) = 13/8 = 1.625.$ This is precisely the square of the ratio of standard deviations reported in the question, $(0.1140/0.08944)^2$.
This analysis has provided a perspective in which the distinction between the two standard deviation calculations in the question can be both understood and used to gain insight into the data-generation mechanism:
The two calculations differ due to possible fluctuations in the subjects' capabilities.
Their ratio (when squared) can be interpreted as an ANOVA F-statistic, permitting its use in evaluating whether the apparent fluctuations may be due to chance or should be accepted as real.
To answer the question, then, one might wish to report both standard deviation calculations together with the F-like statistic given by the square of their ratio.