Reporting standard deviations instead of variances
I think you are right in that standard deviation of each PC can perhaps be a more reasonable or a more intuitive (for some) measure of its "influence" than its variance. And actually it even has a clear mathematical interpretation: variances of PCs are eigenvalues of the covariance matrix, but standard deviations are singular values of the centered data matrix [only scaled by $1/\sqrt{n-1}$].
So yes, it is completely fine to report it. Moreover, e.g. R does report standard deviations of PCs rather than their variances. For example running this simple code:
irispca <- princomp(iris[-5])
summary(irispca)
results in this:
Importance of components:
Comp.1 Comp.2 Comp.3 Comp.4
Standard deviation 2.0494032 0.49097143 0.27872586 0.153870700
Proportion of Variance 0.9246187 0.05306648 0.01710261 0.005212184
Cumulative Proportion 0.9246187 0.97768521 0.99478782 1.000000000
There are standard deviations here, but not variances.
Explained variance
A PC that contains 95% of the data variance might contain only 80% of the variation in the data as measured in standard deviations: isn't the latter a better descriptor?
However, note that after presenting standard deviations, R does not display a "proportion of standard deviation", but instead a proportion of variance. And there is a very good reason for that.
Mathematically, total variance (being a trace of covariance matrix) is preserved under rotations. This means that the sum of variance of original variables is equal to the sum of variances of PCs. In case of the same Fisher Iris dataset, this sum is equal to $4.57$, and so we can say that PC1, having a variance of $2.05^2=4.20$ explains $92\%$ of the total variance.
But the sum of standard deviations is not preserved! The sum of standard deviations of original variables is $3.79$. The sum of standard deviations of PCs is $2.98$. They are not equal! So if you want to say that PC1 with standard deviation $2.05$ explains $x\%$ of the "total standard deviation", what would you take as this total? There is no answer, because it simply does not make sense.
The bottom line is that it is completely fine to look at the standard deviation of each PC and even compare them between each other, but if you want to talk about "explained" something, then only "explained variance" makes sense.
Best Answer
One useful property of the standard deviation is that it has the same units as the mean, so the magnitudes of $\sigma_X$ and $\bar X$ are directly comparable. I've never seen anyone compute the co-standard deviation (by which I assume you mean the square root of the covariance); if the units of $X$ and $Y$ are denoted as $[X]$ and $[Y]$, then the units of the covariance are $[X][Y]$ and the units of the co-standard deviation would be $\sqrt{[X][Y]}$, which isn't particularly useful (unless $X$ and $Y$ have the same units). On the other hand, the correlation $\sigma_{XY}/(\sigma_X \sigma_Y)$ is unitless, and is a very common scale for reporting associations.
The variance (in contrast to the standard deviation) is useful because it generally has nicer mathematical properties; in particular
$$ \sigma^2_{X+Y} = \sigma^2_X + \sigma^2_Y + 2 \sigma_{XY}, $$ which simplifies nicely when $X$ and $Y$ are independent (hence $\sigma_{XY}=0$).
While you're thinking about ways of scaling variances you could also consider the coefficient of variation $\sigma_X/\bar X$ (which is unitless), or the variance-to-mean ratio $\sigma^2_X/\bar X$ (which has weird units but is meaningful in the context of a count distribution such as the Poisson, which is also unitless).