Yes, you can overfit logistic regression models. But first, I'd like to address the point about the AUC (Area Under the Receiver Operating Characteristic Curve):
There are no universal rules of thumb with the AUC, ever ever ever.
What the AUC is is the probability that a randomly sampled positive (or case) will have a higher marker value than a negative (or control) because the AUC is mathematically equivalent to the U statistic.
What the AUC is not is a standardized measure of predictive accuracy. Highly deterministic events can have single predictor AUCs of 95% or higher (such as in controlled mechatronics, robotics, or optics), some complex multivariable logistic risk prediction models have AUCs of 64% or lower such as breast cancer risk prediction, and those are respectably high levels of predictive accuracy.
A sensible AUC value, as with a power analysis, is prespecified by gathering knowledge of the background and aims of a study apriori. The doctor/engineer describes what they want, and you, the statistician, resolve on a target AUC value for your predictive model. Then begins the investigation.
It is indeed possible to overfit a logistic regression model. Aside from linear dependence (if the model matrix is of deficient rank), you can also have perfect concordance, or that is the plot of fitted values against Y perfectly discriminates cases and controls. In that case, your parameters have not converged but simply reside somewhere on the boundary space that gives a likelihood of $\infty$. Sometimes, however, the AUC is 1 by random chance alone.
There's another type of bias that arises from adding too many predictors to the model, and that's small sample bias. In general, the log odds ratios of a logistic regression model tend toward a biased factor of $2\beta$ because of non-collapsibility of the odds ratio and zero cell counts. In inference, this is handled using conditional logistic regression to control for confounding and precision variables in stratified analyses. However, in prediction, you're SooL. There is no generalizable prediction when you have $p \gg n \pi(1-\pi)$, ($\pi = \mbox{Prob}(Y=1)$) because you're guaranteed to have modeled the "data" and not the "trend" at that point. High dimensional (large $p$) prediction of binary outcomes is better done with machine learning methods. Understanding linear discriminant analysis, partial least squares, nearest neighbor prediction, boosting, and random forests would be a very good place to start.
An unbalanced data set (like you suggest when you say the proportion of households with CHE = 1 make up 3% of your data) will influence the results of a logistic regression. See the answers to this question for instance.
In short, the class imbalance in your data set affects only the intercept term. If you are using your model to say something about you4 8 socioeconomic variables, you should be fine. If it is for prediction, then you could consider lowering your probability threshold below 0.5.
Best Answer
A logistic regression doesn't "agree" with anything because the nature of the outcome is 0/1 and the nature of the prediction is a continuous probability. Agreement requires comparable scales: 0.999 does not equal 1. One way of developing a classifier from a probability is by dichotomizing at a threshold. The obvious limitation with that approach: the threshold is arbitrary and can be artificially chosen to produce very high or very low sensitivity (or specificity). Thus, the ROC considers all possible thresholds.
A discriminating model is capable of ranking people in terms of their risk. The predicted risk from the model could be way off, but if you want to design a substudy or clinical trial to recruit "high risk" participants, such a model gives you a way forward. Preventative tamoxifen is recommended for women in the highest risk category of breast cancer as the result of such a study.
Discrimination != Calibration. If my model assigns all non-events a probability of 0.45 and all events a probability of 0.46, the discrimination is perfect, even if the incidence/prevalence is <0.001.