No, that's not right. That's only true if switching capacitance and frequency is the same. The formula for dynamic power dissipation is fCV^2. If I have two parts at the same voltage and one draws more current, obviously the one drawing more current draws more power (P=VI). The current for a device like a RAM is due to charging and discharging of MOS capacitors, hence the first equation. (I designed microprocessors for 15 years and lived this stuff every day).
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And also not true, unless all else is equal. Very efficient memory (that runs at low current) may be hotter due to all sorts of material and packaging effects. Some packages dissipate more heat per square centimeter than other packages, so a device that runs at 1 volt may be hotter than a device that runs at 1.2 volts. And there is no correlation between power use and temperature (again, unless everything else is identical).