Can a Turing machine decide the language $L_\emptyset$ of machines with empty language?
By marking, you probably mean reachability analysis - looking for a path from the initial state to an accepting state. Indeed, the language of a DFA is empty iff there is no such path.

Let's start with an example as to why this fails in TMs. Consider a TM that, in $q_0$, ignores it's input, but writes $a$ on it's tape, moves the head right and goes to state $q_1$, then in $q_1$ it again ignores the input, writes $a$, moves the head left and goes to $q_2$. In $q_2$, if it reads $a$, then it writes $a$, moves the head right and goes back to $q_1$.

That is, the machine just writes $a$ and alternates between two states ($q_1$ and $q_2$) and always has two adjacent $a$'s on the tape. 

We now add a transition from $q_2$ that when reading $b$ goes to an accepting state and halts.

The language of this machine is empty. Indeed, the run always gets stuck in the $q_1-q_2$ loop, and will never get to the accepting state. Yet, there is a state-path to an accepting state. So what went wrong?

Well, intuitively, the ``state'' of a TM is not informative enough to describe the continuation of the run. In order to have all the information, you need the configuration of the TM, which includes the state, the position of the head, and the contents of the tape. If you find a configuration-path (which is called a run) to an accepting configuration, then indeed the language is non-empty, and it is an iff condition. 

The problem with using reachability analysis on the configuration graph, is that it may be infinite. This is why deciding language emptiness is undecidable. 

This is also why language non-emptiness is recognizable - you can perform a BFS on the infinite configuration graph. If there is a path to an accepting state, you will find it eventually. If there isn't, however, then you may get stuck in an infinite search.