How to get rid of voltage drop of this timer circuit?
I tested the regulator circuit with 12VAC 1kHz from a reliable source
  and I got 6V and 1A on 6R load perfectly.


I believe you. If your reliable source happens to have sufficient current compliance then I don't see why it wouldn't work.


  But when I try using the timer circuit above, which supplies the same
  12VAC 1kHz(without load), voltage drops occur and screw up everything.


Yeah. Because of \$R_9\$ in your timer circuit. As long as you are not pulling any current through \$R_9\$, it won't drop any voltage and you will see an unloaded output voltage. The problem comes in trying to supply a load through \$R_9\$.


  D1 diode is just blocking negative half cycle, I can't get rid of it,
  can I?


No, you need it or something to provide that kind of service for you. I wasn't suggesting that you remove it. I was trying to guess why you see anything at all when you load your \$6\:\textrm{V}\$ output with anything much.



Let's back off for a moment and focus on your LM317 regulator circuit. Without the intended load of hundreds of milliamps, and just as you show the circuit itself without a load present, you are already loading the input source to it. You are, because you have a voltage-setting resistor pair present: \$R_1\$ and \$R_2\$. The current in these is:

$$I=\frac{1.25\:\textrm{V}}{180\:\Omega}\approx 7\:\textrm{mA}$$

This is an average current load that exists without any external load on your regulator. And it must be supplied by your input source supply to it. This comes from your capacitor, \$C_1\$. Let's assume for a moment that you need about \$3\:\textrm{V}\$ above the output voltage in order to operate the LM317 properly, for now. This means the capacitor minimum voltage should be about \$9\:\textrm{V}\$. Let's work out the impact on your top circuit.

Since we know the average current required just to maintain the output voltage of the LM317 and have an estimate of the input voltage needed, as well, we can compute the energy (U is sometimes used, so is W, depending on various conventions in use and the field -- I'll use U here) per cycle (at \$1\:\textrm{kHz}\$) is:

$$U = \frac{9\:\textrm{V}\:\cdot\: 7\:\textrm{mA}}{1\:\textrm{kHz}}=63\:\mu\textrm{J}$$

This energy comes from the capacitor and must be replaced each cycle. We can work out the voltage difference needed on the capacitor in the following way:

$$\begin{align*}
U &amp;=\frac{1}{2}\: C\: V^2\\\\
\textrm{d} U &amp;= C\: V\:\textrm{d}V\\\\
\textrm{d}V&amp;=\frac{\textrm{d} U}{C\: V}=\frac{63\:\mu\textrm{J}}{470\:\mu\textrm{F}\:\cdot\:9\:\textrm{V}}\approx 15\:\textrm{mV}
\end{align*}$$

Now we need to compute the current required to achieve that change. Taking into account the fact that \$D_1\$ limits the period, even with a square wave input [and would be much less time with a sine wave], to less than \$500\:\mu\textrm{s}\$, this is just:

$$\begin{align*}
I_C &amp;= C\:\cdot\frac{\textrm{d} V}{\textrm{d} t}\gt 470\:\mu\textrm{F}\:\cdot\frac{15\:\textrm{mV}}{500\:\mu\textrm{s}}\approx 15\:\textrm{mA}
\end{align*}$$

Let's round this up to about \$20\:\textrm{mA}\$. Across \$R_9\$ this leads to a voltage drop of \$330\:\Omega\:\cdot\: 20\:\textrm{mA}=6.6\:\textrm{V}\$! The diode drop across \$D_1\$ would mean that this is more than \$7\:\textrm{V}\$. And if the capacitor were really able to be at \$9\:\textrm{V}\$ (which it obviously could not be), then the diode would be forced into reverse bias (which assumes that it was forward biased in order to allow the current in the first place.)

In short, it's not going to work. Not even to just maintain the output voltage without an external load. Let alone trying to do that with an external load.