Is the converse of a false conditional always true as in the Truth Table? Accroding to the Truth Table, 

If $p$ is TRUE, and $q$ is FALSE, then $p\implies q$ is FALSE.
And the converse, $q \implies p$, is TRUE.

If the conditional statement is 

"If two angles are congruent, they are not equal."
this is a FALSE statement.

Its converse is 
"If two angles are not equal, they are congruent."
The converse is also FALSE.

Why does this example contradict the Truth Table??
The source of the confusion is that you are not dealing merely with implications in propositional logic, where, as you said, the truth tables ensure that $(p\implies q)\lor(q\implies p)$ is always true.  Your statements about angles are universally quantified statements, even though the English language lets you hide the quantifiers.  Your first statement really means "For every two angles $x$ and $y$, if $x$ and $y$ are congruent then $x$ and $y$ are not equal."  Similarly for the second statement. 

So the logical form of these statements is $(\forall x)(\forall y)\,(P(x,y)\implies Q(x,y))$ and $(\forall x)(\forall y)\,(Q(x,y)\implies P(x,y))$.  Because of the quantifiers, it is entirely possible for both of these to be false.  All that's needed for that to happen is that some particular $x_0$ and $y_0$ satisfy $P$ but not $Q$, while a different pair $x_1$ and $y_1$ satisfy $Q$ but not $P$.

If quantifiers are not involved, for example if you have two particular angles and are making statements about just this pair, not angles in general, then a proposition and its converse will not both be false.
It doesn't. In your example let $q = $"two angles are not equal" be false so that means they are equal. Hence the statement "If two angles are not equal, they are congruent" is vacuously true.