Boost converter and a pulsed load
Using a large smoothing capacitor like C1, isolated from the boost output as with R1, may be the only way to use the LMR62014 for this. The LMR62014 has a switch current limit of about \$2A\$, which means that for \$L_o\$ of \$10 \mu H\$, \$V_{\text{in}}\$ of \$3.3 V\$, and \$f_s\$ of \$1.6 {\text{MHz}}\$, the most current delivered to a \$9 V\$ output will be ~ \$0.65A\$. So, it can't support high current pulses on its own. This may be a blessing, as there would be no need to try and get much bandwidth out of the control loop. Just let the bulk capacitance do the job. 

Of course, C1 of \$470 \mu F\$ may not be enough depending on how much droop you can stand during the pulse. Using the equation provided by ThePhoton, shows that maximum expected droop for a \$300 \mu sec\$, \$2 A\$ pulse would be about \$1.3V\$, for a part with negligible esr.

It will be required to have some kind of load at the output of the boost at all times, otherwise \$V_{\text{out}}\$ will rise until something limits it by breaking down. A load of \$40 {\text{mA}}\$ (\$R_o\$ of \$225 \Omega\$) would keep operation in CCM if \$L_o\$ = \$10 \mu H\$ and \$V_{\text{in}}\$ &lt; \$4.7V\$. 

For best dynamic performance it's undesirable to change conduction modes, which will mean in this case stay in CCM. The CCM/DCM boundary is defined as: 

\$L_{\text{crit}}\$ = \$\frac{(\text{dc}-1)^2 \text{dc } R_o}{2 f_s}\$ ; Where \$\text{dc}\$ = \$\frac{V_{\text{out}}-V_{\text{in}}}{V_{\text{out}}}\$ 

A good reference is "Understanding Boost Power Stages". If you're going to be designing a boost, that's a good place to start.