Can any symplectomorphism (1 Definition of canonical transformation) be represented by the flow of a vectorfield? For this question I will use the definition that a canonical transformation is a map $T(q,p)$ from the phase space onto itself, which leaves the symplectic 2-form invariant (which is the definition of a symplectomorphism). 
I know that if the topology of the phase space is allright (which means there are no holes, and any closed path can be drawn to a point), then any vectorfield $V(q,p)$, with  a flow $F(q, p, \alpha)$ with $\frac{\partial F}{\partial \alpha} = V(q,p)$ and $F(q,p, 0) = (q, p)$ can be generated by a function on the phase space. 
What I'd like to know now is wether every symplectomorphism (not a flow, just a symplectomorphism) $T(q,p)$ can be seen as a flow, that means, for every T, I find a flow $F$ so that $T(q,p) = F(q, p, \alpha)$ for a certain value of $ \alpha $.
For example, take the symplectomorphism $T(q, p) = (p, -q)$.  I could represent this symplectomorpohism with the Flow 
$$
F(q, p, \alpha) = (cos(\alpha) q + sin (\alpha) p , -sin(\alpha) q + cos (\alpha) p) 
$$
Then it would hold that $F(q, p, \frac{\pi}{2}) = T(q, p)$. 
Is that true in general? do I find such a flow for any symplectomorphism I want to look at?  
I couldn't think of a counter example, but I don't know how to show it either.   
 A: OP asks good questions.
First of all, let us mention that there is a bijective correspondence between 1-parameter symplectic flows and parameter-independent$^1$ symplectic vector fields. The latter are by definition vector fields $X\in \Gamma(TM)$ that preserve the symplectic 2-form ${\cal L}_X\omega=0$.
OP is essentially asking the following (some of it in previous versions of the question).

  
*
  
*Is the symplectomorphism group path-connected?
  

Answer: Not necessarily, there could be topological obstructions.
2D counterexample: Let the phase space $M=\mathbb{S}^2$ be the 2-sphere equipped with the standard symplectic 2-form $\omega$. The second homotopy group $\pi_2(\mathbb{S}^2)\cong\mathbb{Z}$ is non-trivial. $\Box$


  
*Is any symplectomorphism the 1-time of a 1-parameter symplectic flow, i.e. is it the exponential $\exp(X)$ of a symplectic vector field $X$?
  

Answer: Not necessarily, there could be topological obstructions.
Conjectured 2D counterexample: Consider the phase space $M=\mathbb{R}^2$ with the symplectic 2-form $\omega =\mathrm{d}p\wedge \mathrm{d}q$. Consider the canonical transformation
$$ \begin{bmatrix} Q \cr P \end{bmatrix}
~=~ A\begin{bmatrix} q \cr p \end{bmatrix}, \qquad 
A~:=~\begin{bmatrix} -2 & 1\cr 3 & -2  \end{bmatrix}~\in~Sp(2,\mathbb{R})~=~SL(2,\mathbb{R}). $$ 
We claim that this symplectomorphism can not be generated by a symplectic flow. A tell-tale fact is that the matrix $A$ has no square root. 
Another clue is that this symplectomorphism only has the origin as a fixed point. This means that a symplectic vector field $X$ (for a flow, if it exists) can at most vanish at the origin.
Interestingly, one can write this symplectomorphism as a canonical transformation with a type-2 generating function
$$F_2(q,P)~=~-\frac{1}{2}qP +\frac{3}{4}q^2 - \frac{1}{4}P^2. $$
However we conjecture that a 1-parameter deformation from the identity $F_2(q,P)=qP$ must always go through a singular point. 
See also this related Math.SE post. $\Box$


  
*Is any symplectic vector field $X\in \Gamma(TM)$ a Hamiltonian vector field?
  

Answer: Not necessarily, there could be topological obstructions. In fact, this is measured by the first Poisson cohomology group.
2D counterexample: Consider the phase space $M=\mathbb{R}^2\backslash\{(0,0)\}$ with the symplectic 2-form $\omega =\mathrm{d}p\wedge \mathrm{d}q$. One may check that the vector field
$$X=\frac{q}{q^2+p^2}\frac{\partial}{\partial q} +\frac{p}{q^2+p^2}\frac{\partial}{\partial p} $$
is symplectic but it is not a Hamiltonian vector field. The problem is that the candidate ${\rm arg}(q+ip)$ for the Hamiltonian generator is multi-valued, and hence not globally well-defined. 
See also e.g. this & this related Phys.SE posts. $\Box$
--
$^1$ Note that there exists a notion of flow corresponding to a parameter-dependent vector field. We shall not consider this in this answer. Such flows are used in e.g. this related MO.SE post.
