No this doesn't hold. Here is a simple counter example:
Assume P(X<=a) >= E(X)/a
then P(X<=a) = 1-exp(ka)
and E(X) = 1/k
but 1-exp(ka) >= 1/ka doesn't hold.
To get a feeling for why the Markov inequality can't be reversed I suggest looking up how the proof of the original inequality works. Essentially it relies on the fact that I(X>=a)a <= X where I is the indicator function. But I(X<=a)a >= X does not hold.