Given more information, can a probability lessen? Let $A$, $B$ and $C$ be events in the same probability space. Does 
$$\begin{align}
\mathbb P(A\,|\,B\cup C) \ge \mathbb P(A\,|\,B)
\end{align}$$
hold?
 A: It doesn't. Let $X$ be the outcome of a die roll. 
$P(X=2|X \text{ is even})=\frac{1}{3}$
$P(X=2|X \text{ is even or odd})=\frac{1}{6}$
However, this isn't a case of gaining more information. I'm fairly certain $H(X|B)\leq H(X|B \cup C)$ where $H$ denotes Shannon information (a measure of information/uncertainty). 
A: Assume $P(B)>0$ to avoid division by zero in the argument below.
Because $B \subseteq B\cup C$, we see that 
\begin{equation}P(A\cap B)\leq P(A\cap (B\cup C))\,.\end{equation} On the other hand, $$P(A\cap B)=P(B)P(A\,|\,B)$$
$$P(A\cap (B\cup C))=P(B\cup C)P(A\,|\,(B\cup C))\,,$$ so that
$$P(B)P(A\,|\,B)\leq P(B\cup C)P(A\,|\,(B\cup C))$$ or equivalently
$$P(A\,|\,B)\leq\frac{P(B\cup C)}{P(B)}P(A\,|\,(B\cup C))\,.$$
In general, the fraction on the right hand side can be any number which is at least one, which shows that everything is possible in the general case, as others have remarked. Only in case $P(B\cup C)=P(B)$ do we get that $P(A\,|\,B)\leq P(A\,|\,(B\cup C))\,.$ 
Edit: Do see @whuber's reminder below regarding the fact that $P(B)>0$ is a real restriction.
A: Let me recursively use Bayes theorem for this: 
$$\begin{align}P(A|B \cup C)&=\frac{P(B\cup C|A)P(A)}{P(B\cup C)}\\
           &=\frac{P(A|B)P(B)}{P(B\cup C)}+\frac{P(C|A)P(A)}{P(B\cup C)}\end{align}$$
But we know that $P(B)\leq P(B \cup C)$ and equality only when $C=\emptyset$. So if $C$ is empty or a set with  probability measure $0$ then we can always make the statement $P(A|B \cup C)\geq P(A|B)$. Otherwise it is difficult to compare.
A: The answer to your question in the title is yes.  More information can cause the probability to lessen.
The statement in the post is not valid, but is satisfieable.
In general;
$$
\mathbb P(A\,|\,B\cap C) > \mathbb P(A\,|\,B) \iff P(A\cap B\,|\,C)>P(A\cap B)\\
\mathbb P(A\,|\,B\cap C) = \mathbb P(A\,|\,B) \iff P(A\cap B\,|\,C)=P(A\cap B)\\
\mathbb P(A\,|\,B\cap C) < \mathbb P(A\,|\,B) \iff P(A\cap B\,|\,C)<P(A\cap B)
$$
For a concrete example consider a deck of card with all heart suits values less than 10 removed.  If B is "a card is drawn" and A is "An Ace is drawn" how will the probability of A behaive if you are given the information that a heart is drawn? Or if you get the information that a heart is not drawn?
