Yes, deterministic context-free languages are closed under union with regular languages.
It is easy to show they are closed under intersection with regular languages. We can apply a product construction. When reading a symbol we also do a step of a FSA that is running in parallel. Accepting a computation means accepting both the DPDA part and the FSA part.
This doesn't work for union. Why? Because a DPDA may end up in an infinite loop of $\varepsilon$-transitions, so it will never read all of its input. In that way we cannot test acceptance by the FSA.
In order to make this work we also need a normal form for DPDA, that they will never do such infinite loops, and always read all input. That normal form is usually proved when showing DCFL are closed under complement (and is non-trivial).
By he way, once we know that DCFL is closed under complement and closed under intersection with regular languages, closure under union with regular languages follows by De Morgan: $L\cup R = (L^c \cap R^c)^c$.